\(ĐKXĐ:x\ne\pm1\)
\(A=\frac{\left(\frac{x+1}{x-1}-\frac{x^2+x}{x^2-1}\right)}{\left(\frac{1}{x+1}+\frac{1}{x-1}\right)}=\frac{\frac{x+1}{\left(x-1\right)}-\frac{x}{x-1}}{\frac{x-1}{\left(x+1\right)\left(x-1\right)}+\frac{x+1}{\left(x+1\right)\left(x-1\right)}}=\frac{\frac{x+1-x}{x-1}}{\frac{x-1+x+1}{\left(x+1\right)\left(x-1\right)}}\\ =\frac{1}{x-1}\cdot\frac{\left(x+1\right)\left(x-1\right)}{2x}=\frac{x+1}{2x}\)
\(\left|x-2\right|=1\\ \Rightarrow\left[{}\begin{matrix}x-2=1\left(x\ge2\right)\\-x+2=1\left(x< 2\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\left(TM\right)\)
Tại x = 3 ( TM )
\(\Rightarrow A=\frac{3+1}{2}=\frac{4}{2}=2\)
Tại x = 1 ( KTM ) => Giá trị biếu thức A không xác định tại x = 1
