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 \(x-2011=0\)

\(x=2011\)

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

\(\Leftrightarrow\left(\frac{x+4}{2011}+1\right)+\left(\frac{x+3}{2012}+1\right)-\left(\frac{x+2}{2013}+1\right)-\left(\frac{x+1}{2014}+1\right)=0\)

\(\Leftrightarrow\frac{x+2015}{2011}+\frac{x+2015}{2012}-\frac{x+2015}{2013}-\frac{x+2015}{2014}=0\)

\(\Leftrightarrow\left(x+2015\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)

\(\Leftrightarrow x+2015=0\) (Vì: \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\) )

\(\Leftrightarrow x=-2015\)

bạn nói kết quả không thôi ư mình cần cả cách giải

Bạn xem lại đề nhé U Suck

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

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

\(\frac{x+2015}{2014}+\frac{x+2015}{2013}=\frac{x+2015}{2012}+\frac{x+2015}{2011}\)

\(\frac{x+2015}{2014}+\frac{x+2015}{2013}-\frac{x+2015}{2012}-\frac{x+2015}{2011}=0\)

\(\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)=0\)

Vì \(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\ne0\) nên x + 2015 = 0

x = 0 - 2015

x = -2015

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

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

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

\(\frac{x+2015}{2014}+\frac{x+2015}{2013}=\frac{x+2015}{2012}+\frac{x+2015}{2011}\)

\(\Rightarrow\frac{x+2015}{2014}+\frac{x+2015}{2013}-\frac{x+2015}{2012}-\frac{x+2015}{2011}=0\)

\(\Rightarrow\left(x+2015\right)\left(\frac{1}{2014}+\frac{1}{2013}-\frac{1}{2012}-\frac{1}{2011}\right)=0\)

=> x + 2015 = 0 ( vì 1/2014 + 1/2013 - 1/2012  - 1/2011 khác 0)

=> x = -2015

x = - 2015

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

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

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

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

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

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

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 !

\(\frac{x-1}{2011}+\frac{x-1}{2012}+\frac{x-1}{2013}=\frac{x-1}{2014}+\frac{x-1}{2015}\)

\(\Rightarrow\frac{x-1}{2011}+\frac{x-1}{2012}+\frac{x-1}{2013}-\frac{x-1}{2014}-\frac{x-1}{2015}=0\)

\(\left(x-1\right).\left(\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\right)=0\)

mà \(\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}-\frac{1}{2015}\ne0\)

=> x - 1 = 0

x = 1

bn có chép sai đề ko z???

a)  \(\frac{x-90}{10}+\frac{x-76}{12}+\frac{x-58}{14}+\frac{x-36}{16}+\frac{x-15}{17}=15\)

\(\Leftrightarrow\frac{x-90}{10}-1+\frac{x-76}{12}-2+\frac{x-58}{14}-3+\frac{x-36}{16}-4+\frac{x-1}{17}-5=0\)

\(\Leftrightarrow\frac{x-90-10}{10}+\frac{x-76-2.12}{12}+\frac{x-58-3.14}{14}+\frac{x-36-4.16}{16}+\frac{x-15-5.17}{17}=0\)

\(\Leftrightarrow\frac{x-100}{10}+\frac{x-100}{12}+\frac{x-100}{14}+\frac{x-100}{16}+\frac{x-100}{17}=0\)

\(\Leftrightarrow\left(x-100\right)\left(\frac{1}{10}+\frac{1}{12}+\frac{1}{14}+\frac{1}{16}+\frac{1}{17}\right)=0\)

         \(\Leftrightarrow x-100=0\Leftrightarrow x=100\)

Vậy  \(S=\left\{100\right\}\)

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

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

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

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

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

        \(\Leftrightarrow x+4024=0\Leftrightarrow x=-4024\)

Vậy  \(S=\left\{-4024\right\}\)

Phương trình a bạn trừ phân thức đầu tiên cho 1, phân thức thứ hai cho 2, phân thức thứ ba cho 3, phân thức thứ tư cho 4, phân thức thứ năm cho 5, vế còn lại trừ đi 15. Tiếp theo bạn đặt x -100 làm nhân tử chung. Cuối cùng tìm được x= 100

Phương trình b bạn chỉ cần chuyển vế sang rồi giải tương tự phương trình a kết quả là x=-4024

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

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

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

\(\frac{x+2015}{2011}+\frac{x+2015}{2012}=\frac{x+2015}{2013}+\frac{x+2015}{2014}\)

\(\frac{x+2015}{2011}+\frac{x+2015}{2012}-\frac{x+2015}{2013}-\frac{x+2015}{2014}=0\)

\(\left(x+2015\right)\left(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\right)=0\)

vì \(\frac{1}{2011}+\frac{1}{2012}-\frac{1}{2013}-\frac{1}{2014}\ne0\)nên:

x+2015=0

x=-2015