$\frac{\left(.\right)\left(.\right)}{=3=}$bn thấy hình trên giống j

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Hãy tham gia nhóm Học sinh Hoc24OLM

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Gửi câu hỏi ẩn danh

Nguyễn Dương Nguyệt Linh

17 tháng 4 2019 lúc 21:04

$\frac{\left(.\right)\left(.\right)}{=3=}$

bn thấy hình trên giống j

Lớp 1 Toán Câu hỏi của OLM

lutufine 159732486

17 tháng 4 2019 lúc 21:04

nó giống cái kính bị hỏng và dang sủa

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Khắc lên đây ba dòng thơ...

17 tháng 4 2019 lúc 21:05

giống ĐÔRAEMON

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๖ۣۜmạnͥh2ͣkͫ5ツ

17 tháng 4 2019 lúc 21:05

giống mặt đoremon

hok tốt

nhớ tk

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Thiên tài thiện xạ

17 tháng 4 2019 lúc 21:05

mon ú đúng ko

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Bé's Bông's

17 tháng 4 2019 lúc 21:05

trả lời:

Theo mk giống doraemon

đoán thế

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♡๖ۣۜTɦàηɦ Đạт๖ۣۜ ²ƙ⁸ ♡

17 tháng 4 2019 lúc 21:07

doraemon đấy

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Ngo Nguyen Minh Ngoc

17 tháng 4 2019 lúc 21:08

doraemon

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☠️🔥{{A©E}}🔥☠️

17 tháng 4 2019 lúc 21:08

mon nhưng thiếu dâu

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Kato kid

17 tháng 4 2019 lúc 21:15

giống Doraemon

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Nấm

17 tháng 4 2019 lúc 22:48

Doraemon nhưng ko giống

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♚мưα❖︵(🅳🅴🆅🅸🅻)๖︵²⁰⁰⁵

23 tháng 4 2019 lúc 18:50

I. Nội qui tham gia "Giúp tôi giải toán"

1. Không đưa câu hỏi linh tinh lên diễn đàn, chỉ đưa các bài mà mình không giải được hoặc các câu hỏi hay lên diễn đàn;

2. Không trả lời linh tinh, không phù hợp với nội dung câu hỏi trên diễn đàn.

3. Không "Đúng" vào các câu trả lời linh tinh nhằm gian lận điểm hỏi đáp.

Các bạn vi phạm 3 điều trên sẽ bị giáo viên của Online Math trừ hết điểm hỏi đáp, có thể bị khóa tài khoản hoặc bị cấm vĩnh viễn không đăng nhập vào trang web.

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Các câu hỏi tương tự

manh

15 tháng 3 2019 lúc 12:53

$frac{45}{53}.left(frac{17}{3}-frac{53}{47}right)+frac{17}{3}.left(frac{6}{17}-frac{47}{53}right)$$frac{45}{53}.left(frac{17}{3}-frac{53}{47}right)+frac{17}{3}.left(frac{6}{17}-frac{47}{53}right)$$frac{45}{53}.left(frac{17}{3}-frac{53}{47}right)+frac{17}{3}.left(frac{6}{17}-frac{47}{53}right)$$frac{45}{53}.left(frac{17}{3}-frac{53}{47}right)+frac{17}{3}.left(frac{6}{17}-frac{47}{53}right)$$frac{45}{53}.left(frac{17}{3}-frac{53}{47}right)+frac{17}{3}.left(frac{6}{17}-frac{47}{53}right)$$frac{45}{...

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$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}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17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$$\frac{45}{53}.\left(\frac{17}{3}-\frac{53}{47}\right)+\frac{17}{3}.\left(\frac{6}{17}-\frac{47}{53}\right)$v

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Lớp 1 Toán Câu hỏi của OLM

Nguyễn Đăng Khôi

17 tháng 3 2020 lúc 21:04

$\frac{x\left(x+1\right)}{2\left(x-3\right)\left(x+1\right)}+\frac{x\left(x-3\right)}{2\left(x-3\right)\left(x+1\right)}-\frac{4x}{2\left(x-3\right)\left(x+1\right)}=0$

$\frac{x^2+x+x^2-3x-4x}{2\left(x-3\right)\left(x+1\right)}=0$

$x^2-3x=0$

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Lớp 1 Toán Câu hỏi của OLM

pham thi thu trang

18 tháng 4 2018 lúc 20:52

b, MA-Bfrac{sqrt{x}+2}{sqrt{x}+3}-left(frac{5}{x+sqrt{x}-6}+frac{1}{sqrt{x}-2}right)frac{sqrt{x}+2}{sqrt{x}+3}-frac{5}{x+sqrt{x}-6}-frac{1}{sqrt{x}-2}frac{left(sqrt{x}+2right)left(sqrt{x}-2right)}{x+sqrt{x}-6}-frac{5}{x+sqrt{x}-6}-frac{1left(sqrt{x}+3right)}{x+sqrt{x}-6}frac{x-4-5-sqrt{x}-3}{left(sqrt{x}+3right)left(sqrt{x}-2right)}frac{x-sqrt{x}-12}{left(sqrt{x}+3right)left(sqrt{x}-2right)}frac{x-4sqrt{x}+3sqrt{x}-12}{left(sqrt{x}+3right)left(sqrt{x}-2right)}frac{left(sqrt{x}-4right)left(sqrt{x...

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b, $M=A-B=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\left(\frac{5}{x+\sqrt{x}-6}+\frac{1}{\sqrt{x}-2}\right)$

$=\frac{\sqrt{x}+2}{\sqrt{x}+3}-\frac{5}{x+\sqrt{x}-6}-\frac{1}{\sqrt{x}-2}$

$=\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}{x+\sqrt{x}-6}-\frac{5}{x+\sqrt{x}-6}-\frac{1\left(\sqrt{x}+3\right)}{x+\sqrt{x}-6}$

$=\frac{x-4-5-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\frac{x-4\sqrt{x}+3\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}$$=\frac{\left(\sqrt{x}-4\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}-4}{\sqrt{x}-2}$

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Lớp 1 Toán Câu hỏi của OLM

Tuấn

28 tháng 12 2015 lúc 22:13

ĐKXD: tự làm nhé từ đề Rightarrow Afrac{2sqrt{y}-9}{left(sqrt{y}-2right)left(sqrt{y}-3right)}-frac{left(sqrt{y}+3right)left(sqrt{y}-3right)}{left(sqrt{y}-2right)left(sqrt{y}-3right)}+frac{left(2sqrt{y}+1right)left(sqrt{y}-2right)}{left(sqrt{y}-2right)left(sqrt{y}-3right)}Rightarrow Afrac{2sqrt{y}-9-y+9+2y-3sqrt{y}-2}{MC}Rightarrow Afrac{y-sqrt{y}-2}{MC}frac{left(sqrt{y}-2right)left(sqrt{y}+1right)}{left(sqrt{y}-2right)left(sqrt{y}-3right)}frac{sqrt{y}+1}{sqrt{y}-3}Đc nhé bác :DSensodai: ĐỀ NGHỊ...

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ĐKXD: tự làm nhé =='
từ đề
$\Rightarrow A=\frac{2\sqrt{y}-9}{\left(\sqrt{y}-2\right)\left(\sqrt{y}-3\right)}-\frac{\left(\sqrt{y}+3\right)\left(\sqrt{y}-3\right)}{\left(\sqrt{y}-2\right)\left(\sqrt{y}-3\right)}+\frac{\left(2\sqrt{y}+1\right)\left(\sqrt{y}-2\right)}{\left(\sqrt{y}-2\right)\left(\sqrt{y}-3\right)}$
$\Rightarrow A=\frac{2\sqrt{y}-9-y+9+2y-3\sqrt{y}-2}{MC}$
$\Rightarrow A=\frac{y-\sqrt{y}-2}{MC}=\frac{\left(\sqrt{y}-2\right)\left(\sqrt{y}+1\right)}{\left(\sqrt{y}-2\right)\left(\sqrt{y}-3\right)}=\frac{\sqrt{y}+1}{\sqrt{y}-3}$
Đc nhé bác :D
Sensodai: ĐỀ NGHỊ CÁC THÀNH PHẦN TAY NHANH HƠN NÃO K CMT NHÉ =='

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Lớp 1 Toán Câu hỏi của OLM

Phùng Minh Quân

8 tháng 8 2019 lúc 18:27

sqrt{a}+sqrt{b}lesqrt{2left(a+bright)}VPsqrt{ab}left(sqrt{a}+sqrt{b}right)lefrac{a+b}{2}sqrt{2left(a+bright)}RightarrowVP^2lefrac{left(a+bright)^3}{2} (1) chứng minh bổ đề: VT^2left(frac{left(a+bright)^2}{2}+frac{a+b}{4}right)^2gefrac{left(a+bright)^3}{2}Leftrightarrowfrac{left(a+bright)^4}{4}+frac{left(a+bright)^2}{16}+frac{left(a+bright)^3}{4}gefrac{left(a+bright)^3}{2}Leftrightarrowleft(a+bright)^4+frac{left(a+bright)^2}{4}geleft(a+bright)^3Có: left(a+bright)^4+frac{left(a+bright)^2}{4}ge2sqr...

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$\sqrt{a}+\sqrt{b}\le\sqrt{2\left(a+b\right)}$

$VP=\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)\le\frac{a+b}{2}\sqrt{2\left(a+b\right)}$$\Rightarrow$$VP^2\le\frac{\left(a+b\right)^3}{2}$ (1)

chứng minh bổ đề: $VT^2=\left(\frac{\left(a+b\right)^2}{2}+\frac{a+b}{4}\right)^2\ge\frac{\left(a+b\right)^3}{2}$

$\Leftrightarrow$$\frac{\left(a+b\right)^4}{4}+\frac{\left(a+b\right)^2}{16}+\frac{\left(a+b\right)^3}{4}\ge\frac{\left(a+b\right)^3}{2}$

$\Leftrightarrow$$\left(a+b\right)^4+\frac{\left(a+b\right)^2}{4}\ge\left(a+b\right)^3$

Có: $\left(a+b\right)^4+\frac{\left(a+b\right)^2}{4}\ge2\sqrt{\frac{\left(a+b\right)^6}{4}}=\left(a+b\right)^3$$\Rightarrow$$VT^2\ge\frac{\left(a+b\right)^3}{2}$ (2)

(1) và (2) => $VT^2\ge VP^2$ => $VT\ge VP$ ( đpcm )

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Lớp 1 Toán Câu hỏi của OLM

Nguyễn Khang

8 tháng 9 2019 lúc 20:36

Khi thử đổi biến chứng minh Iran 96 và cái kết.... Mà chả biết lúc đổi biến có tính sai chỗ nào ko mà kết quả nó nhìn khủng khiếp quá:(Cho a, b, c là các số không âm thỏa mãn không có 2 số nào đồng thời bằng 0. Chứng minh rằng:left(ab+bc+caright)left(frac{1}{left(a+bright)^2}+frac{1}{left(b+cright)^2}+frac{1}{left(c+aright)^2}right)gefrac{9}{4}Đặt left(a+b+c;ab+bc+ca;abcright)left(3u;3v^2;w^3right)Cần chứng minhleft(ab+bc+caright)left(frac{1}{left(a+bright)^2}+frac{1}{left(b+cright)^2}+frac{1}{l...

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Khi thử đổi biến chứng minh Iran 96 và cái kết.... Mà chả biết lúc đổi biến có tính sai chỗ nào ko mà kết quả nó nhìn khủng khiếp quá:(

Cho a, b, c là các số không âm thỏa mãn không có 2 số nào đồng thời bằng 0. Chứng minh rằng:

$\left(ab+bc+ca\right)\left(\frac{1}{\left(a+b\right)^2}+\frac{1}{\left(b+c\right)^2}+\frac{1}{\left(c+a\right)^2}\right)\ge\frac{9}{4}$

Đặt $\left(a+b+c;ab+bc+ca;abc\right)=\left(3u;3v^2;w^3\right)$

Cần chứng minh

$\left(ab+bc+ca\right)\left(\frac{1}{\left(a+b\right)^2}+\frac{1}{\left(b+c\right)^2}+\frac{1}{\left(c+a\right)^2}\right)\ge\frac{9}{4}$

$\Leftrightarrow v^2\left(\left(3v^2+a^2\right)^2+\left(3v^2+b^2\right)^2+\left(3v^2+c^2\right)^2\right)\ge3\left(9uv^2-w^3\right)$

$\Leftrightarrow v^2\left(27v^4+6v^2\left(a^2+b^2+c^2\right)+a^4+b^4+c^4\right)\ge3\left(9uv^2-w^3\right)$

$\Leftrightarrow v^2\left(27v^4+6v^2\left(9u^2-6v^2\right)+a^4+b^4+c^4\right)\ge3\left(9uv^2-w^3\right)$

$\Leftrightarrow v^2\left(27v^4+6v^2\left(9u^2-6v^2\right)+81u^4-108u^2v^2+18v^4+12uw^3\right)\ge3\left(9uv^2-w^3\right)$

$\Leftrightarrow135u^4v^2-144u^2v^4+12uv^2w^3-27uv^2+45v^6+3w^3\ge0$

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Lớp 1 Toán Câu hỏi của OLM

Incursion_03

26 tháng 5 2019 lúc 10:10

Đã bảo là liên hợp là ra mà đ tin hả Zũ ? -_- x^3+sqrt{left(x+1right)^3}9x+8left(xge-1right)Leftrightarrowleft(x^3+1right)+left(x+1right)sqrt{x+1}-9left(x+1right)0Leftrightarrowleft(x+1right)left(x^2-x+1+sqrt{x+1}-9right)0Leftrightarroworbr{begin{cases}x-1left(Tmright)x^2-x+sqrt{x+1}-80left(1right)end{cases}}Giải left(1right)Leftrightarrowleft(x^2-3xright)+left(2x-6right)+left(sqrt{x+1}-2right)0 Leftrightarrow xleft(x-3right)+2left(x-3right)+frac{x-3}{sqrt{x+1}+2}0 ...

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Đã bảo là liên hợp là ra mà đ tin hả Zũ ? -_-

$x^3+\sqrt{\left(x+1\right)^3}=9x+8\left(x\ge-1\right)$

$\Leftrightarrow\left(x^3+1\right)+\left(x+1\right)\sqrt{x+1}-9\left(x+1\right)=0$

$\Leftrightarrow\left(x+1\right)\left(x^2-x+1+\sqrt{x+1}-9\right)=0$

$\Leftrightarrow\orbr{\begin{cases}x=-1\left(Tm\right)\\x^2-x+\sqrt{x+1}-8=0\left(1\right)\end{cases}}$

Giải $\left(1\right)\Leftrightarrow\left(x^2-3x\right)+\left(2x-6\right)+\left(\sqrt{x+1}-2\right)=0$

$\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)+\frac{x-3}{\sqrt{x+1}+2}=0$

$\Leftrightarrow\left(x-3\right)\left(x+2+\frac{1}{\sqrt{x+1}+2}\right)=0$

Vì x > -1 nên dễ thấy cái ngoặc to > 0

Do đó x = 3

Vậy có 2 nghiệm -1 và 3 (nghiệm thứ 3 nào nữa nhỉ ? -,-'' )

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Lớp 1 Toán Câu hỏi của OLM

Chicken Dinner

24 tháng 8 2021 lúc 21:29

1. frac{x^3-10x^2+25x}{x^2-5x}0 ( đkxđ: xne0;5) frac{xleft(x-5right)^2}{xleft(x-5right)}0 x-50 vô no2. Afrac{2x^2-2}{x^3-x^2-4x+4}frac{2left(x-1right)left(x+1right)}{left(x-1right)left(x-2right)left(x+2right)} ( a, đkxđ: xne1;pm2)b, A0 frac{2left(x+1right)}{left(x-2right)left(x+2right)}0 x-1( TM) . Vậy A0Leftrightarrow x-13. Bfrac{3x^2-12}{left(x-3right)left(x^2+4x+4right)}frac{3left(x-2right)left(x+2right)}{left(x-3right)left(x+2right)^2} ( a, đkxđ: xne3,-2)b,B0Leftrightarrow...

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1. $\frac{x^3-10x^2+25x}{x^2-5x}$$=0$ ( đkxđ: $x\ne0;5$)

<=> $\frac{x\left(x-5\right)^2}{x\left(x-5\right)}=0$<=> $x-5=0$<=> vô n_o

_{2. $A=$$\frac{2x^2-2}{x^3-x^2-4x+4}$$=\frac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x-2\right)\left(x+2\right)}$} ( a, đkxđ: $x\ne1;\pm2$)

b, $A=0$<=> $\frac{2\left(x+1\right)}{\left(x-2\right)\left(x+2\right)}=0$<=> $x=-1$( TM) . Vậy $A=0\Leftrightarrow x=-1$

3. $B=\frac{3x^2-12}{\left(x-3\right)\left(x^2+4x+4\right)}$$=\frac{3\left(x-2\right)\left(x+2\right)}{\left(x-3\right)\left(x+2\right)^2}$ ( a, đkxđ: $x\ne3,-2$)

$b,B=0\Leftrightarrow\frac{3\left(x-2\right)}{\left(x-3\right)\left(x+2\right)}=0\Leftrightarrow x=2\left(tm\right)$. Vậy $B=0\Leftrightarrow x=2$

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Lớp 1 Toán Câu hỏi của OLM

Nguyễn Khang

18 tháng 9 2019 lúc 20:39

@Ai đó:vTìm min của 2x^2 + y^2 +z^2 biết xy + yz + zx 1 và x, y, z 0Cách của em như sau(ko chắc đâu nhé, cách này em mới nghĩ ra thôi): Ta cho k 0thỏa mãn Age kleft(xy+yz+zxright)Hay2x^2-xleft(ky+kzright)+y^2-kyz+z^2ge0Có:VT2left(x-frac{ky+kz}{4}right)^2+frac{left(8-k^2right)y^2-left(2k^2+8kright)yz+left(8-k^2right)z^2}{8}2left(x-frac{ky+kz}{4}right)^2+frac{left(8-k^2right)left(y-frac{left(2k^2+8zright)z}{2left(8-k^2right)}right)^2+frac{z^2}{4}left[4left(8-k^2right)-frac{left(2k^2+8kright)^2}{...

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@Ai đó:v

Tìm min của 2x^2 + y^2 +z^2 biết xy + yz + zx = 1 và x, y, z > 0

Cách của em như sau(ko chắc đâu nhé, cách này em mới nghĩ ra thôi): Ta cho k >0thỏa mãn $A\ge k\left(xy+yz+zx\right)$

Hay

$2x^2-x\left(ky+kz\right)+y^2-kyz+z^2\ge0$

Có:$VT=2\left(x-\frac{ky+kz}{4}\right)^2+\frac{\left(8-k^2\right)y^2-\left(2k^2+8k\right)yz+\left(8-k^2\right)z^2}{8}$

$=2\left(x-\frac{ky+kz}{4}\right)^2+\frac{\left(8-k^2\right)\left(y-\frac{\left(2k^2+8z\right)z}{2\left(8-k^2\right)}\right)^2+\frac{z^2}{4}\left[4\left(8-k^2\right)-\frac{\left(2k^2+8k\right)^2}{8-k^2}\right]}{8}$

Bây giờ để bđt là luôn đúng thì $8-k^2\ge0$ và $4\left(8-k^2\right)=\frac{\left(2k^2+8k\right)^2}{8-k^2}$

Ngay lập tức ta thấy $k=\sqrt{5}-1$

Từ đó..

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Lớp 1 Toán Câu hỏi của OLM

Khoá học trên OLM (olm.vn)

Khoá học trên OLM (olm.vn)