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Em là fan cuồng soái ca...
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soyeon_Tiểu bàng giải
17 tháng 8 2016 lúc 18:41

\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+...+\frac{1}{4900}\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{50}\right)\)

\(=\frac{1}{2}.\frac{49}{50}=\frac{49}{100}\)

Nguyễn Quỳnh Anh BTS
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LÊ TIẾN ĐẠT
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Phạm Tuấn Đạt
8 tháng 7 2018 lúc 21:00

\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)

\(=\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{7}\right)\)

\(=\frac{1}{2}.\frac{6}{7}=\frac{3}{7}\)

Kiên-Messi-8A-Boy2k6
8 tháng 7 2018 lúc 21:05

Đặt \(C=\frac{1}{2}+\frac{1}{12}+\frac{1}{24}+...+\frac{1}{84}\)

\(\Rightarrow\frac{C}{2}=1+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{42}\)

\(\Rightarrow C.\frac{1}{2}=1+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{6.7}\)

\(\Rightarrow C.\frac{1}{2}=1+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{6}-\frac{1}{7}\)

\(\Rightarrow C.\frac{1}{2}=1+\frac{1}{2}-\frac{1}{7}\)

\(\Rightarrow C=\left(1+\frac{1}{2}-\frac{1}{7}\right).2\)

nene
8 tháng 7 2018 lúc 21:06

\(\frac{3}{7}\)

Trần Tích Thường
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Fudo
11 tháng 7 2018 lúc 12:53

Tính nhanh : 

\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)

\(A=2\left(\frac{1}{2\cdot4}+\frac{1}{4\cdot6}+\frac{1}{6\cdot8}+\frac{1}{8\cdot10}+\frac{1}{10\cdot12}+\frac{1}{12\cdot14}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\right)\)

\(A=2\left(\frac{1}{2}-\frac{1}{14}\right)\)

\(A=2\cdot\frac{3}{7}\)

\(A=\frac{6}{7}\)

Nguyễn Thanh Hiền
11 tháng 7 2018 lúc 12:54

\(A=\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}\)

\(A=\frac{2}{8}+\frac{2}{24}+\frac{2}{48}+\frac{2}{80}+\frac{2}{120}+\frac{2}{168}\)

\(A=\frac{2}{2.4}+\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+\frac{2}{10.12}+\frac{2}{12.14}\)

\(A=\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}\)

\(A=\frac{1}{2}-\frac{1}{14}\)

\(A=\frac{3}{7}\)

_Chúc bạn học tốt_

TAKASA
11 tháng 7 2018 lúc 13:09

A=1/4+1/12+1/24+1/40+1/60+1/84

A=2/8+2/24+2/48+2/80+2/120+2/168

A=1/2-1/4+1/4-1/6+1/6-1/8+1/8-1/10+1/10-1/12+1/12-1/14

A=1/2-1/14

A=7/14-1/14

A=6/14

A=3/7

Chi Nguyễn
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soyeon_Tiểu bàng giải
16 tháng 8 2016 lúc 11:10

\(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+...+\frac{1}{4900}\)

\(=\frac{1}{2}.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{2450}\right)\)

\(=\frac{1}{2}.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{49.50}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{50}\right)\)

\(=\frac{1}{2}.\frac{49}{50}=\frac{49}{100}\)

Hảo Đào thị mỹ
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Minh Hiền Trần
20 tháng 5 2016 lúc 8:16

\(\frac{1}{2x^2+10x+12}+\frac{1}{2x^2+14x+24}+\frac{1}{2x^2+18x+40}+\frac{1}{2x^2+22x+60}=\frac{1}{8}\)

<=> \(\frac{1}{2x^2+6x+4x+12}+\frac{1}{2x^2+6x+8x+24}+\frac{1}{2x^2+8x+10x+40}+\frac{1}{2x^2+12x+10x+60}=\frac{1}{8}\)

<=> \(\frac{1}{2x\left(x+3\right)+4\left(x+3\right)}+\frac{1}{2x\left(x+3\right)+8\left(x+3\right)}+\frac{1}{2x\left(x+4\right)+10\left(x+4\right)}+\frac{1}{2x\left(x+6\right)+10\left(x+6\right)}=\frac{1}{8}\)

<=> \(\frac{1}{\left(x+3\right)\left(2x+4\right)}+\frac{1}{\left(x+3\right)\left(2x+8\right)}+\frac{1}{\left(x+4\right)\left(2x+10\right)}+\frac{1}{\left(x+6\right)\left(2x+10\right)}=\frac{1}{8}\)

<=> \(\frac{1}{2\left(x+2\right)\left(x+3\right)}+\frac{1}{2\left(x+3\right)\left(x+4\right)}+\frac{1}{2\left(x+4\right)\left(x+5\right)}+\frac{1}{2\left(x+5\right)\left(x+6\right)}=\frac{1}{8}\)

<=> \(\frac{1}{2}.\left[\frac{1}{\left(x+2\right)\left(x+3\right)}+\frac{1}{\left(x+3\right)\left(x+4\right)}+\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}\right]=\frac{1}{8}\)

<=> \(\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}+\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}=\frac{1}{8}:\frac{1}{2}\)

<=> \(\frac{1}{x+2}-\frac{1}{x+6}=\frac{1}{4}\)

<=> \(\frac{4\left(x+6\right)-4\left(x+2\right)}{4\left(x+2\right)\left(x+6\right)}=\frac{\left(x+2\right)\left(x+6\right)}{4\left(x+2\right)\left(x+6\right)}\)

<=> \(4\left(x+6\right)-4\left(x+2\right)=\left(x+2\right)\left(x+6\right)\)

<=> \(4\left(x+6-x-2\right)=x^2+8x+12\)

<=> \(4.4=x^2+8x+12\)

<=> \(x^2+8x-4=0\)

<=> ...

Đến đây bạn tự giải tiếp. Mình bấm máy 570ES PLUS II thì ra nghiệm \(x\approx0,47\).

 

 

QQQWWW
20 tháng 5 2016 lúc 6:16

icon-chat

Nguyen tuan cuong
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tuancl
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Hoàng Thanh Huyền
20 tháng 10 2019 lúc 7:55

Ta có: \(B=\frac{1}{112}-\frac{1}{84}-\frac{1}{60}-\frac{1}{40}-\frac{1}{24}-\frac{1}{12}-\frac{1}{4}\)

\(\Rightarrow2B=\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{20}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)

\(\Rightarrow2B=\frac{1}{56}-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}\right)\)

\(\Rightarrow2B=\frac{1}{56}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}\right)\)

\(\Rightarrow2B=\frac{1}{56}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}\right)\)

\(\Rightarrow2B=\frac{1}{56}-\left(1-\frac{1}{7}\right)\)

\(\Rightarrow2B=\frac{1}{56}-\frac{6}{7}\)

\(\Rightarrow2B=-\frac{47}{56}\)

\(\Rightarrow B=-\frac{47}{112}\)

Hok tốt nha^^

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xamcon
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