a: ĐKXĐ: \(x\notin\left\{1;-1;-\dfrac{1}{2}\right\}\)
b: \(N=\left(\dfrac{2}{x+1}-\dfrac{1}{x-1}-\dfrac{5}{1-x^2}\right):\dfrac{2x+1}{x^2-1}\)
\(=\left(\dfrac{2}{x+1}-\dfrac{1}{x-1}+\dfrac{5}{\left(x-1\right)\left(x+1\right)}\right)\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(=\dfrac{2\left(x-1\right)-x-1+5}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{\left(x-1\right)\left(x+1\right)}{2x+1}\)
\(=\dfrac{2x-2-x+4}{2x+1}=\dfrac{x+2}{2x+1}\)
c: N=3
=>\(x+2=3\left(2x+1\right)\)
=>6x+3=x+2
=>5x=-1
=>\(x=-\dfrac{1}{5}\left(nhận\right)\)
d: Để N nguyên thì \(x+2⋮2x+1\)
=>\(2x+4⋮2x+1\)
=>\(2x+1+3⋮2x+1\)
=>\(3⋮2x+1\)
=>\(2x+1\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{0;-1;1;-2\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{0;-2\right\}\)
