a)
Biểu thức được xác định khi
\(\left\{{}\begin{matrix}2x-2\ne0\\x^2-1\ne0\\2x+2\ne0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x\ne1\\x\ne\pm1\\x\ne-1\end{matrix}\right.\)
b)
\(\left(\dfrac{x+1}{2x-2}+\dfrac{3}{x^2-1}-\dfrac{x+3}{2x+2}\right).\dfrac{4x^2-4}{5}\)
\(=\left(\dfrac{x+1}{2\left(x-1\right)}+\dfrac{3}{\left(x-1\right)\left(x+1\right)}-\dfrac{x+3}{2\left(x+1\right)}\right).\dfrac{\left(2x-2\right)\left(2x+2\right)}{5}\)
\(=\dfrac{\left(x+1\right)\left(x+1\right)+3.2-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}.\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{x^2+2x+1+6-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}.\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{10}{2\left(x-1\right)\left(x+1\right)}.\dfrac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=4\)
