a) Ta có: \(A=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}-\dfrac{1}{1-x}+\dfrac{2-x^2}{x^2-x}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}\)
\(=\dfrac{x^2}{x-1}\)
b) Ta có: |2x-5|=3
\(\Leftrightarrow\left[{}\begin{matrix}2x-5=-3\\2x-5=3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=2\\2x=8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(loại\right)\\x=4\left(nhận\right)\end{matrix}\right.\)
Thay x=4 vào A, ta được:
\(A=\dfrac{4^2}{4-1}=\dfrac{16}{3}\)