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

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

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

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

Vì \(\left(\frac{1}{2009}+\frac{1}{2010}-\frac{1}{2011}-\frac{1}{2012}\right)\ne0\)

Nên biểu thức (1) xảy ra khi \(x+2013=0\)

\(x=-2013\)

b) \(\left(x-2011\right)\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)  (2)

Vì \(\left(\frac{1}{2010}+\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)\ne0\)

Nên biểu thức (2) xảy ra khi \(x-2011=0\)

\(x=2011\)

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

Ta có \(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)

\(\Rightarrow\)\(1+\dfrac{x+4}{2010}+1+\dfrac{x+3}{2011}=1+\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}\)

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

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

\(\Rightarrow\)x+2014=0

\(\Rightarrow x=0-2014=-2014\)

Vậy x=-2014

\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)

\(\Leftrightarrow\dfrac{x+4}{2010}+1+\dfrac{x+3}{2011}+1=\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}+1\)

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

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

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

\(\Leftrightarrow x+2014=0\).Do \(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\ne0\)

\(\Leftrightarrow x=-2014\)

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

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

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

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

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

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

\(\Leftrightarrow x=2012\)

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

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

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

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

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

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 ?

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

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

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

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

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

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

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 !

Ta có:\(\frac{x-1}{2013}+\frac{x-2}{2012}=\frac{x-3}{2011}+\frac{x-4}{2010}\)

\(\Rightarrow\left(\frac{x-1}{2013}-1\right)+\left(\frac{x-2}{2012}-1\right)=\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-4}{2010}-1\right)\)

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

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

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

Vì \(\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\ne0\)nên để biểu thức =0 

\(\Leftrightarrow x-2014=0\Rightarrow x=2014\)

Ta cTa có: 2013 x − 1 + 2012 x − 2 = 2011 x − 3 + 2010 x − 4 ⇒ 2013 x − 1 − 1 + 2012 x − 2 − 1 = 2011 x − 3 − 1 + 2010 x − 4 − 1 ⇒ 2013 x − 2014 + 2012 x − 2014 = 2011 x − 2014 + 2010 x − 2014 ⇒ 2013 x − 2014 + 2012 x − 2014 − 2011 x − 2014 − 2010 x − 2014 = 0 ⇒ x − 2014 . 2013 1 + 2012 1 − 2011 1 − 2010 1 = 0 
1 = 0

chúc bn hok tốt @_@

TA CÓ: \(\frac{x}{2013}+\frac{x-1}{2012}=\frac{x-2}{2011}+\frac{x-3}{2010}\)

\(\Rightarrow\frac{x}{2013}+\frac{x-1}{2012}-\frac{x-2}{2011}-\frac{x-3}{2010}=0\)

\(\frac{x}{2013}-1+\frac{x-1}{2012}-1-\frac{x-2}{2011}+1-\frac{x-3}{2010}+1=0\)

\(\left(\frac{x}{2013}-1\right)+\left(\frac{x-1}{2012}-1\right)-\left(\frac{x-2}{2011}-1\right)-\left(\frac{x-3}{2010}-1\right)=0\)

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

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

MÀ \(\frac{1}{2013}< \frac{1}{2011};\frac{1}{2012}< \frac{1}{2010}\)

\(\Rightarrow\frac{1}{2013}+\frac{1}{2012}-\frac{1}{2011}-\frac{1}{2010}\ne0\)

\(\Rightarrow x-2013=0\)

\(x=2013\)

VẬY X= 2013

CHÚC BN NĂM MỚI VUI VẺ NHA!!!!!!!