\(\frac{x-1}{2011}+\frac{x-2}{2012}=\frac{x-3}{2013}+\frac{x-4}{2014}\)

\(\frac{x-1}{2011}+1+\frac{x-2}{2012}+1=\frac{x-3}{2013}+1+\frac{x-4}{2014}+1\)

\(\Rightarrow\frac{x+2010}{2011}+\frac{x+2010}{2012}=\frac{x+2010}{2013}+\frac{x+2010}{2014}\)

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

\(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}>0\)

\(\Leftrightarrow x+2010=0\Rightarrow x=-2010\)

Bạn tiếp tục áp dụng phương pháp này vào bài 2 nha nhưng bài b bạn sẽ trừ 1 ở mỗi thức

\(a)\) \(\frac{x-1}{2011}+\frac{x-2}{2012}=\frac{x-3}{2013}+\frac{x-4}{2014}\)

\(\Leftrightarrow\)\(\left(\frac{x-1}{2011}+1\right)+\left(\frac{x-2}{2012}+1\right)=\left(\frac{x-3}{2013}+1\right)+\left(\frac{x-4}{2014}+1\right)\)

\(\Leftrightarrow\)\(\frac{x-1+2011}{2011}+\frac{x-2+2012}{2012}=\frac{x-3+2013}{2013}+\frac{x-4+2014}{2014}\)

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

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

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

Vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\)

Nên \(x-2010=0\)

\(\Rightarrow\)\(x=2010\)

Vậy \(x=2010\)

Chúc bạn học tốt ~ 

a.\(\left(\frac{x-1}{2011}+1\right)+\left(\frac{x-2}{2012}+1\right)=\left(\frac{x-3}{2013}+1\right)\left(\frac{x-4}{2014}+1\right)\)

=> \(\frac{x+2010}{2011}+\frac{x+2010}{2012}=\frac{x+2010}{2013}+\frac{x+2010}{2014}\)

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

=>x+2010=0 (vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\)\(\ne0\)

=> x = -2010

Vậy x= -2010

Câu b làm tương tự nhưng -1 nhé !

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

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

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}=0\end{cases}}\)

mà \(\frac{1}{2009}+\frac{1}{2010}+\frac{1}{2011}-\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\)

nên \(x+1=0\)

\(\Leftrightarrow x=-1\)

Bài 1 : 

Ta có  : 

\(\frac{x+2011}{2013}+\frac{x+2012}{2012}=\frac{x+2010}{2014}+\frac{x+2013}{2011}\)

\(\Rightarrow\left(\frac{x+2011}{2013}+1\right)+\left(\frac{x+2012}{2012}+1\right)=\left(\frac{x+2010}{2014}+1\right)\)

\(+\left(\frac{x+2013}{2011}+1\right)\)

\(\Rightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}=\frac{x+4024}{2014}+\frac{x+4024}{2011}\)

\(\Rightarrow\frac{x+4024}{2013}+\frac{x+4024}{2012}-\frac{x+4024}{2014}-\frac{x+4024}{2011}=0\)

\(\Rightarrow\left(x+4024\right)\left(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2014}-\frac{1}{2011}\right)=0\)

\(\Rightarrow x+4024=0\)

\(\Rightarrow x=-4024\)

Bài 2 : 

Đặt \(x^2+2x+1=a\Rightarrow a=\left(x+1\right)^2\ge0\)

=> Phương trình trở thành 

\(\frac{a}{a+1}+\frac{a+1}{a+2}=\frac{7}{6}\)

\(\Rightarrow\frac{a}{a+1}.6\left(a+1\right)\left(a+2\right)+\frac{a+1}{a+2}.6\left(a+1\right)\left(a+2\right)=\frac{7}{6}.6\left(a+1\right)\left(a+2\right)\)

\(\Rightarrow6a\left(a+2\right)+6\left(a+1\right)^2=7\left(a+1\right)\left(a+2\right)\)

\(\Rightarrow12a^2+24a+6=7a^2+21a+14\)

\(\Rightarrow5a^2+3a-8=0\)

\(\Rightarrow\left(a-1\right)\left(5a+8\right)=0\)

Vì \(a\ge0\Rightarrow a=1\)

\(\Rightarrow x^2+2x+1=1\)

\(x^2+2x=0\)

\(\Rightarrow x\left(x+2\right)=0\)

\(\Rightarrow x\in\left\{-2,0\right\}\)

a)

\(2^x\left(1+2+2^2+2^3\right)=480\)

\(2^x.15=480\Rightarrow2^x=\frac{480}{15}=32=2^5\Rightarrow x=5\)

Chính Xác 100% là X=5 

k cho mink nhé các pạn

$\frac{\frac{2010}{2011}}{\frac{2012}{2013}}+\frac{\frac{2011}{2012}}{\frac{2013}{2014}}+\frac{\frac{2012}{2013}}{\frac{2014}{2015}}$

$\frac{\frac{2010}{2011}}{\frac{2012}{2013}}+\frac{\frac{2011}{2012}}{\frac{2013}{2014}}+\frac{\frac{2012}{2013}}{\frac{2014}{2015}}$

$\frac{\frac{2010+2011+2012}{2011+2012+2013}}{\frac{2012+2013+2014}{2013+2014+2015}}$

$\frac{\frac{2010}{2011}+\frac{2011}{2012}+\frac{2012}{2013}}{\frac{2012+2013+2014}{2013+2014+2015}}$

$\frac{\frac{2010+2011+2012}{2011+2012+2013}}{\frac{2012}{2013}+\frac{2013}{2014}+\frac{2014}{2015}}$

dễ ợt nhưng éo biết làm thông cảm nha

ban Dang Ha Trang an noi gi ki vay 

Áp dụng BĐT Cô - si ngược dấu :

\(\sqrt{x-2010}=\frac{1}{2}\sqrt{4\left(x-2010\right)}\le\frac{4+\left(x-2010\right)}{4}\)

\(\Rightarrow\sqrt{x-2010}-1\le\frac{4+\left(x-2010\right)}{4}-1=\frac{x-2010}{4}\)

\(\Rightarrow\frac{\sqrt{x-2010}-1}{x-2010}\le\frac{1}{4}\)

Hoàn toàn tương tự với những phân thức còn lại 

\(\Rightarrow\frac{\sqrt{x-2010}-1}{x-2010}+\frac{\sqrt{y-2011}-1}{y-2011}\le\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x-2010=4\\x-2011=4\\z-2012=4\end{cases}\Leftrightarrow\hept{\begin{cases}x=2014\\y=2015\\z=2016\end{cases}}}\)

\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)

\(\Rightarrow2^x\cdot1+2^x\cdot2^1+2^x\cdot2^2+2^x\cdot2^3=480\)

\(\Rightarrow2^x\left(1+2^1+2^2+2^3\right)=480\)

\(\Rightarrow2^x\cdot15=480\)

\(\Rightarrow2^x=32\Rightarrow2^x=2^5\Rightarrow x=5\)

b) \(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\frac{2012}{1}+\frac{2011}{2}+...+\frac{2}{2011}+\frac{1}{2012}\)

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\left(\frac{2011}{2}+1\right)+...+\left(\frac{2}{2011}+1\right)+\left(\frac{1}{2012}+1\right)+1\)

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=\frac{2013}{2}+...+\frac{2013}{2011}+\frac{2013}{2012}+\frac{2013}{2013}\)

\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)x=2013\left(\frac{1}{2}+...+\frac{1}{2012}+\frac{1}{2013}\right)\)

\(\Rightarrow x=2013.\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}}\)

\(\Rightarrow x=2013\)

Vậy \(x=2013\)

đề này ở đâu đấy bạn ?

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

\(\Leftrightarrow\frac{x+1}{2012}+1+\frac{x+2}{2011}+1=\frac{x+3}{2010}+1+\frac{x+4}{2009}+1\)

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

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

\(\frac{x+1}{2012}+\frac{X+2}{2011}=\frac{X+3}{2010}+\frac{X+4}{2009}.\)

\(\Leftrightarrow\frac{X+1}{2012}+\frac{X+2}{2011}+2=\frac{X+3}{2010}+\frac{X+4}{2009}+2\)

\(\Leftrightarrow\frac{x+1}{2012}+1+\frac{x+2}{2011}+1=\frac{x+3}{2010}+1+\frac{x+4}{2009}+1\)

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

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

Mà \(\frac{1}{2012}+\frac{1}{2011}-\frac{1}{2010}-\frac{1}{2009}>0\)

\(\Leftrightarrow x+2013=0\)

\(\Leftrightarrow x=-2013\)

KL ; PT có Nghiệm \(S=\left\{-2013\right\}\)

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

\(\Leftrightarrow\frac{x+1}{2012}+1+\frac{x+2}{2011}+1=\frac{x+3}{2010}+1+\frac{x+4}{2009}+1\)

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

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

Vậy ....