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Duy anh
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⭐Hannie⭐
16 tháng 7 2023 lúc 20:32

\(\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}=-3\\ \Rightarrow\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}+3=0\\ \Rightarrow\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+6}{2004}+1\right)+\left(\dfrac{x+7}{2003}+1\right)=0\\ \Rightarrow\dfrac{x+2010}{2005}+\dfrac{x+2010}{2004}+\dfrac{x+2010}{2003}=0\\ \Rightarrow\left(x+2010\right)\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\right)=0\\ \Rightarrow x+2010=0\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\ne0\right)\\ \Rightarrow x=-2010\)

Cô nàng kì bí
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Feed Là Quyền Công Dân
21 tháng 8 2017 lúc 21:10

ai ra đề cho 1 lạy

Cô nàng kì bí
21 tháng 8 2017 lúc 21:18

HELP ME!

hoc24.vn
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Adonis Baldric
5 tháng 8 2017 lúc 12:41

\(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}=\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-3}{2006}\)

\(\Leftrightarrow\left(\dfrac{x-8}{2001}+1\right)+\left(\dfrac{x-7}{2002}+1\right)+\left(\dfrac{x-6}{2003}+1\right)=\left(\dfrac{x-5}{2004}+1\right)+\left(\dfrac{x-4}{2005}+1\right)+\left(\dfrac{x-3}{2006}+1\right)\)

\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}-\dfrac{x-2009}{2004}-\dfrac{x-2009}{2005}-\dfrac{x-2009}{2006}=0\)

\(\Leftrightarrow\left(x-2009\right).\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)=0\)

\(\text{Mà}:\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)\ne0\)

\(\Rightarrow x-2009=0\Rightarrow x=2009\)

Trần Quốc Lộc
6 tháng 8 2017 lúc 9:58

\(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}=\dfrac{x-5}{2004}+\dfrac{x-4}{4}+\dfrac{x-5}{2006}\)

\(\Leftrightarrow\left(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}\right)-3=\left(\dfrac{x-5}{2004}+\dfrac{x-4}{4}+\dfrac{x-5}{2006}\right)-3\)

\(\Leftrightarrow\left(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}\right)-\left(1+1+1\right)=\left(\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-5}{2006}\right)-\left(1+1+1\right)\)

\(\Leftrightarrow\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}-1-1-1=\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-5}{2006}-1-1-1\)

\(\Leftrightarrow\left(\dfrac{x-8}{2001}-1\right)+\left(\dfrac{x-7}{2002}-1\right)+\left(\dfrac{x-6}{2003}-1\right)=\left(\dfrac{x-5}{2004}-1\right)+\left(\dfrac{x-4}{2005}-1\right)+\left(\dfrac{x-5}{2006}-1\right)\)

\(\)\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}=\dfrac{x-2009}{2004}+\dfrac{x-2009}{2006}+\dfrac{x-2009}{2006}\)

\(\Leftrightarrow\left(\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}\right)-\left(\dfrac{x-2009}{2004}+\dfrac{x-2009}{2006}+\dfrac{x-2009}{2006}\right)=0\)

\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}-\dfrac{x-2009}{2004}-\dfrac{x-2009}{2006}-\dfrac{x-2009}{2006}=0\)

\(\Leftrightarrow\left(x-2009\right)\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)=0\)

\(\Leftrightarrow x-2009=0\)

\(\Leftrightarrow x=2009\)

Vậy \(x=2009\)

Kamato Heiji
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Nguyễn Lê Phước Thịnh
5 tháng 3 2021 lúc 21:44

a) Ta có: \(\dfrac{2x+1}{6}-\dfrac{x-2}{4}=\dfrac{3-2x}{3}-x\)

\(\Leftrightarrow\dfrac{2\left(2x+1\right)}{12}-\dfrac{3\left(x-2\right)}{12}=\dfrac{4\left(3-2x\right)}{12}-\dfrac{12x}{12}\)

\(\Leftrightarrow4x+2-3x+6=12-8x-12x\)

\(\Leftrightarrow x+8-12+20x=0\)

\(\Leftrightarrow21x-4=0\)

\(\Leftrightarrow21x=4\)

\(\Leftrightarrow x=\dfrac{4}{21}\)

Vậy: \(S=\left\{\dfrac{4}{21}\right\}\)

Akai Haruma
5 tháng 3 2021 lúc 22:00

Hình như em viết công thức bị lỗi rồi. Em cần chỉnh sửa lại để được hỗ trợ tốt hơn!

Akai Haruma
5 tháng 3 2021 lúc 23:03

a) 

PT \(\Leftrightarrow \frac{4x+2}{12}-\frac{3x-6}{12}=\frac{12-8x}{12}-\frac{12x}{12}\)

\(\Leftrightarrow 4x+2-3x+6=12-8x-12x\)

\(\Leftrightarrow 21x=4\Leftrightarrow x=\frac{4}{21}\)

b) 

PT \(\Leftrightarrow \frac{30x+15}{20}-\frac{100}{20}-\frac{6x+4}{20}=\frac{24x-12}{20}\)

\(\Leftrightarrow 30x+15-100-6x-4=24x-12\Leftrightarrow -89=-12\) (vô lý)

Vậy pt vô nghiệm.

Trần Lê Nhi
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Trần Lê Nhi
29 tháng 6 2018 lúc 21:20

câu B là \(2^{12}\) nha mấy bn

Hihiii
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Xử Nữ Chính Là Tôi
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Trần Tuấn Minh
14 tháng 10 lúc 21:32

 

????

 

Monkey D .Luffy
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Lightning Farron
28 tháng 3 2017 lúc 11:19

\(\dfrac{x+1}{2004}+\dfrac{x+2}{2003}+\dfrac{x+3}{2002}+35=2^5\)

\(pt\Leftrightarrow\dfrac{x+1}{2004}+\dfrac{x+2}{2003}+\dfrac{x+3}{2002}+3=0\)

\(\Leftrightarrow\dfrac{x+1}{2004}+1+\dfrac{x+2}{2003}+1+\dfrac{x+3}{2002}+1=0\)

\(\Leftrightarrow\dfrac{x+1}{2004}+\dfrac{2004}{2004}+\dfrac{x+2}{2003}+\dfrac{2003}{2003}+\dfrac{x+3}{2002}+\dfrac{2002}{2002}=0\)

\(\Leftrightarrow\dfrac{x+2005}{2004}+\dfrac{x+2005}{2003}+\dfrac{x+2005}{2002}=0\)

\(\Leftrightarrow\left(x+2005\right)\left(\dfrac{1}{2004}+\dfrac{1}{2003}+\dfrac{1}{2002}\right)=0\)

\(\Rightarrow x+2005=0\). Do \(\dfrac{1}{2004}+\dfrac{1}{2003}+\dfrac{1}{2002}\ne0\)

\(\Rightarrow x=-2005\)

Nguyễn Trần Tuấn Anh
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kuroba kaito
13 tháng 4 2018 lúc 13:38

\(\dfrac{x+1}{2008}+\dfrac{x+2}{2007}+\dfrac{x+3}{2006}=\dfrac{x+4}{2005}+\dfrac{x+5}{2004}+\dfrac{x+6}{2003}\)

\(\dfrac{x+1}{2008}+1+\dfrac{x+2}{2007}+1+\dfrac{x+3}{2006}+1=\dfrac{x+4}{2005}+1+\dfrac{x+5}{2004}+1+\dfrac{x+6}{2003}+1\)

\(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}=\dfrac{x+2009}{2005}+\dfrac{x+2009}{2004}+\dfrac{x+2009}{2003}\)

\(\dfrac{x+2009}{2008}+\dfrac{x+2009}{2007}+\dfrac{x+2009}{2006}-\dfrac{x+2009}{2005}-\dfrac{x+2009}{2004}-\dfrac{x+2009}{2003}=0\)

\(\left(x+2009\right)\left(\dfrac{1}{2008}+\dfrac{1}{2007}+\dfrac{1}{2006}-\dfrac{1}{2005}-\dfrac{1}{2004}-\dfrac{1}{2003}\right)=0\)

⇔ x+2009=0

⇔ x=-2009

vậy x=-2009 là nghiệm của pt

Phùng Khánh Linh
13 tháng 4 2018 lúc 17:31

a) ( x2 + x )2 + 4( x2 + x ) = 12

<=> ( x2 + x )2 + 4( x2 + x ) + 4 - 16 = 0

<=> ( x2 + x + 2)2 - 16 = 0

<=> ( x2 + x + 2 + 4)( x2 + x + 2 - 4) = 0

<=> ( x2 + x + 6 )( x2 + x - 2) = 0

Do : x2 + x + 6

= x2 + 2.\(\dfrac{1}{2}x+\dfrac{1}{4}+6-\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{23}{4}\)\(\dfrac{23}{4}\) > 0 ∀x

=> x2 + x - 2 = 0

<=> x2 - x + 2x - 2 = 0

<=> x( x - 1) + 2( x - 1) = 0

<=> ( x - 1)( x + 2 ) = 0

<=> x = 1 hoặc : x = - 2

KL.....

b) Kuroba kaito làm rùi nhé hihi