a: \(A=\dfrac{x^2+2x+1-x^2+2x-1-4}{\left(x-1\right)\left(x+1\right)}=\dfrac{4x-4}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{x+1}\)
b: Ta có: |x-2|=2x-1
\(\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\\left(2x-1-x+2\right)\left(2x-1+x-2\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>=\dfrac{1}{2}\\\left(x+1\right)\left(3x-3\right)=0\end{matrix}\right.\Leftrightarrow x=1\)(loại)
a.\(A=\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}-\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)
\(A=\dfrac{\left(x+1\right)^2-\left(x-1\right)^2-4}{\left(x-1\right)\left(x+1\right)}\)
\(A=\dfrac{x^2+2x+1-x^2+2x-1-4}{\left(x-1\right)\left(x+1\right)}\)
\(A=\dfrac{4x-4}{\left(x-1\right)\left(x+1\right)}=\dfrac{4\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(A=\dfrac{4}{x+1}\)
b.\(\left|2-x\right|=2x-1\)
\(ĐK:x\ge\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2-x=2x-1\\x-2=2x-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-1\left(ktm\right)\end{matrix}\right.\)
Thế \(x=1\) vào A ta được
\(A=\dfrac{4}{1+1}=\dfrac{4}{2}=2\)
Mà \(x\ne\pm1\) => không có giá trị của x thỏa mãn thế vào A khi \(\left|2-x\right|=2x-1\)
