Question

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

\(\)\(\frac{x}{x-1}+\frac{x-1}{x}=2\)

\(\frac{x-1}{x}+\frac{x-2}{x+1}=2\)

Answer 1

https://i.imgur.com/m6v2e2N.jpg

Answer 2

https://i.imgur.com/D5ypDzg.jpg

Answer 3

https://i.imgur.com/zdFtjXC.jpg

Answer 4

a)ĐK: \(x\ne\pm1\)

pt\(\Leftrightarrow\dfrac{x^2-x-2x^2-2x+3x+3-2x-3}{x^2-1}=0\)

\(\Rightarrow-x^2-2x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)(TM)

Vậy pt có tập nghiệm là S={0;-2}.

b) ĐK: \(x\ne0;1\)

\(\Leftrightarrow\dfrac{x^2+x^2-x-x+1}{x\left(x-1\right)}=2\)

\(\Rightarrow2x^2-2x+1=2x^2-2x\)

\(\Leftrightarrow1=0\left(vl\right)\)

c) ĐK: \(x\ne0;-1\)

pt\(\Rightarrow\left(x-1\right)\left(x+1\right)+x\left(x-2\right)=2x^2+2x\)

\(\Leftrightarrow2x^2-1-2x=2x^2+2x\)

\(\Leftrightarrow4x=-1\)

\(\Leftrightarrow x=\dfrac{-1}{4}\left(TM\right)\)

Vậy pt có tập nghiệm là S=\(\left\{\dfrac{-1}{4}\right\}\)