a)\(=\dfrac{x^2+2xy+y^2}{4\left(x+y\right)}\\ =\dfrac{\left(x+y\right)^2}{4\left(x+y\right)}\\ =\dfrac{x+y}{4}\)

b)\(=\dfrac{x+5}{2x-2}-\dfrac{\left(x-1\right)\left(x+1\right)}{4}.\dfrac{2}{x+1}\\ =\dfrac{x+5}{2\left(x-1\right)}-\dfrac{x-1}{2}\\ =\dfrac{x+5}{2\left(x-1\right)}-\dfrac{x^2-2x+1}{2\left(x-1\right)}\\ =\dfrac{x+5-x^2+2x-1}{2\left(x-1\right)}\\ =\dfrac{-x^2+3x+4}{2\left(x-1\right)}\)

\(\Leftrightarrow\left(3x-5\right)^2-2^2=0\\ \Leftrightarrow\left(3x-5-2\right)\left(3x-5+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x-5-2=0\\3x-5+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=1\end{matrix}\right.\)

Vậy...

\(=4x^2-4x-\left(4x^2-4x+1\right)\\ =4x^2-4x-4x^2+4x+1\\ =1\)

phá ngoặc ra ~~