Lời giải:
a)
ĐKXĐ: \(\left\{\begin{matrix} x^2-7x+10\neq 0\\ x^2-4\neq 0\\ 2-x\neq 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} (x-2)(x-5)\neq 0\\ (x-2)(x+2)\neq 0\\ 2-x\neq 0\end{matrix}\right.\Leftrightarrow x\neq 2; x\neq 5; x\neq -2\)
\(N=\frac{2(x-5)}{(x-2)(x-5)}-\frac{2x}{(x-2)(x+2)}-\frac{1}{x-2}\)
\(=\frac{2}{x-2}-\frac{2x}{(x-2)(x+2)}-\frac{1}{x-2}=\frac{1}{x-2}-\frac{2x}{(x-2)(x+2)}\)
\(=\frac{x+2}{(x-2)(x+2)}-\frac{2x}{(x-2)(x+2)}=\frac{2-x}{(x+2)(x-2)}=\frac{-1}{x+2}\)
b)
Để $N$ nhận giá trị nguyên thì $\frac{-1}{x+2}$ nguyên
$\Rightarrow -1\vdots x+2$
$\Rightarrow x+2\in\left\{\pm 1\right\}$
$\Rightarrow x\in\left\{-3; -1\right\}$ (đều thỏa mãn)