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14 tháng 3 2019

\(\frac{x-1}{2013}+\frac{x-2}{2012}+\frac{x-3}{2011}=\frac{x-4}{2010}+\frac{x-5}{2009}+\frac{x-6}{2008}\)

\(\Leftrightarrow\)\(\left(\frac{x-1}{2013}-1\right)+\left(\frac{x-2}{2012}-1\right)+\left(\frac{x-3}{2011}-1\right)=\left(\frac{x-4}{2010}-1\right)+\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-6}{2008}-1\right)\)

\(\Leftrightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}+\frac{x-2013}{2011}=\frac{x-2014}{2010}+\frac{x-2014}{2009}+\frac{x-2014}{2008}\)

\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)

tự làm nốt~

14 tháng 3 2019

kudo shinichi làm sai ở chỗ:

\(\frac{x-2013}{2011}\)phải là \(\frac{x-2014}{2011}\)mới đúng nhé

6 tháng 2 2018

Ta có :

\(\frac{x+1}{2012}+\frac{x+2}{2011}+\frac{x+3}{2010}=\frac{x+4}{2009}+\frac{x+5}{2008}+\frac{x+6}{2007}\)

\(\left(\frac{x+1}{2012}+1\right)+\left(\frac{x+2}{2011}+1\right)+\left(\frac{x+3}{2010}+1\right)=\left(\frac{x+4}{2009}+1\right)+\left(\frac{x+5}{2008}+1\right)+\left(\frac{x+6}{2007}+1\right)\)

\(\Leftrightarrow\)\(\frac{x+2013}{2012}+\frac{x+2013}{2011}+\frac{x+2013}{2010}=\frac{x+2013}{2009}+\frac{x+2013}{2008}+\frac{x+2013}{2007}\)

\(\Leftrightarrow\)\(\left(x+2013\right).\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}\right)=\left(x+2013\right).\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\right)\)

\(\Leftrightarrow\)\(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}=\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\)\(\left(1\right)\)

Mà \(\frac{1}{2012}< \frac{1}{2009}\)\(;\)\(\frac{1}{2011}< \frac{1}{2008}\)\(;\)\(\frac{1}{2010}< \frac{1}{2007}\)

\(\Rightarrow\)\(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}< \frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}\)\(\left(2\right)\)

Từ \(\left(1\right)\)và \(\left(2\right)\)suy ra không có giá trị nào của \(x\)thoả mãn đề bài 

Vậy không có gía trị nào của \(x\)hay \(x\in\left\{\varnothing\right\}\)

8 tháng 5 2020

\(\frac{x-1}{2013}+\frac{x-2}{2012}+\frac{x-3}{2011}=\frac{x-4}{2010}+\frac{x-5}{2009}+\frac{x-6}{2008}\)  ( có lẽ đề như này ) 

\(\Leftrightarrow\frac{x-1}{2013}-1+\frac{x-2}{2012}-1+\frac{x-3}{2011}-1=\frac{x-4}{2010}-1+\frac{x-5}{2009}-1+\frac{x-6}{2008}-1\)

\(\Leftrightarrow\frac{x-2014}{2013}+\frac{x-2014}{2012}+\frac{x-2014}{2011}-\frac{x-2014}{2010}-\frac{x-2014}{2009}-\frac{x-2014}{2008}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\right)=0\)

\(\Leftrightarrow x-2014=0\left(\frac{1}{2013}+\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}\ne0\right)\)

\(\Leftrightarrow x=2014\)

...

Ta có : \(x^2+9x+20=x^2+4x+5x+20=\left(x+4\right)\left(x+5\right)\)

\(x^2+11x+30=x^2+5x+6x+30=\left(x+5\right)\left(x+6\right)\)

\(x^2+13x+42=x^2+6x+7x+42=\left(x+6\right)\left(x+7\right)\)

\(\Rightarrow Pt\Leftrightarrow\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\) (*)\(ĐKXĐ:x\ne-4;x\ne-5;x\ne-6;x\ne-7\)

(*) \(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\)

\(\Leftrightarrow\frac{x+7-x-4}{\left(x+4\right)\left(x+7\right)}=\frac{1}{18}\)

\(\Leftrightarrow3.18=x^2+4x+7x+28\)

\(\Leftrightarrow x^2-2x+13x-26=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+13\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+13=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=-13\left(tm\right)\end{cases}}}\)

25 tháng 3 2018

\(\frac{x+1}{2011}+\frac{x+2}{2010}=\frac{x+3}{2009}+\frac{x+4}{2008}\Leftrightarrow\frac{x+1}{2011}+1+\frac{x+2}{2010}+1=\frac{x+3}{2009}+1+\frac{x+4}{2008}+1\)

\(\Leftrightarrow\frac{x+1}{2011}+\frac{2011}{2011}+\frac{x+2}{2010}+\frac{2010}{2010}=\frac{x+3}{2009}+\frac{2009}{2009}+\frac{x+4}{2008}+\frac{2008}{2008}\)

\(\Leftrightarrow\frac{x+1+2011}{2011}+\frac{x+2+2010}{2010}=\frac{x+3+2009}{2009}+\frac{x+4+2008}{2008}\)

\(\Leftrightarrow\frac{x+2012}{2011}+\frac{x+2012}{2010}=\frac{x+2012}{2009}+\frac{x+2012}{2008}\)

\(\Leftrightarrow\left(x+2012\right)\left(\frac{1}{2011}+\frac{1}{2010}\right)=\left(x+2012\right)\left(\frac{1}{2009}+\frac{1}{2008}\right)\)

\(\Leftrightarrow\left(x+2012\right)\left(\frac{1}{2011}+\frac{1}{2010}-\frac{1}{2009}-\frac{1}{2008}=0\right)\) 

mà 1/2011+1/2010-1/2009-1/2008 khác 0 

\(\Rightarrow x+2012=0\Rightarrow x=-2012\)

\(\left(3x-2\right)^2-x\left(9x-2\right)=24\Leftrightarrow9x^2-12x+4-9x^2+2x=24\)

\(\Leftrightarrow-10x+4=24\Leftrightarrow-10x=20\Leftrightarrow x=-2\)

25 tháng 3 2018

1;  Ta có : x+1/2011 + x+2/2010 = x+3/2009 + x+4/ 2008

Suy ra: 2+(x+1/2011 + x+2/2010 ) = 2+( x+3/2009 + x+4/2008)

suy ra ban tach 2=1+1 roi cong 1 voi  tưng phân số  trên nha  sẽ ra kết quả ngay thôi 

2; gợi ý nè : (3x-2)^2 =(3x)^2 + 2*3x*2+2^2

7 tháng 3 2017

\(\frac{x-2}{2012}+\frac{x-3}{2011}+\frac{x-4}{2010}+\frac{x-2029}{5}=0\)

\(\Leftrightarrow\frac{x-2}{2012}-1+\frac{x-3}{2011}-1+\frac{x-4}{2010}-1+\frac{x-2029}{5}+3=0\)

\(\Leftrightarrow\frac{x-2014}{2012}+\frac{x-2014}{2011}+\frac{x-2014}{2010}+\frac{x-2014}{5}=0\)

\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+\frac{1}{5}\right)=0\)

\(\Leftrightarrow x-2014=0\).Do \(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+\frac{1}{5}\ne0\)

\(\Leftrightarrow x=2014\)

28 tháng 4 2018

\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}+...+\frac{x+2010}{1}=\left(-2010\right)\)

\(\Rightarrow\left(\frac{x+1}{2010}+1\right)+\left(\frac{x+2}{2009}+1\right)+...+\left(\frac{x+2010}{1}+1\right)=-2010+2010\)

\(\Rightarrow\frac{x+2011}{2010}+\frac{x+2011}{2009}+...+\frac{x+2011}{1}=0\)

\(\Rightarrow\left(x+2011\right)\left(1+\frac{1}{2}+...+\frac{1}{2009}+\frac{1}{2010}\right)=0\)

\(\Rightarrow x+2011=0\Leftrightarrow x=-2011\)

28 tháng 4 2018

x=-2011

7 tháng 1 2018

Phương trình đã cho tương đương với :

\(\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+\frac{x-3}{2010}-1+...+\frac{x-2012}{1}-1+2012=2012\)

\(\Leftrightarrow\)\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+\frac{x-2013}{2010}+...+\frac{x-2013}{1}=0\)

\(\Leftrightarrow\)\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+\frac{1}{2010}+...+\frac{1}{1}\right)=0\)

Tìm x theo như toán lớp 6 nha

\(x-2013=0\)

\(\Leftrightarrow\)\(x=2013\)

7 tháng 1 2018

ta có pt 

<=>\(\frac{x-1}{2012}-1+\frac{x-2}{2011}-1+...+\frac{x-2012}{1}-1=0\)

<=>\(\frac{x-2013}{2012}+\frac{x-2013}{2011}+...+\frac{x-2013}{1}=0\)

<=>\(\left(x-2013\right)\left(\frac{1}{2012}+\frac{1}{2011}+...+\frac{1}{1}\right)=0\Leftrightarrow x-2013=0\Leftrightarrow x=2013\)

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