a: B=1
=>\(\dfrac{3}{x^2-1}=1\)
=>\(x^2-1=3\)
=>\(x^2=4\)
=>\(\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
b: \(A=\left(\dfrac{x+2}{x+1}-\dfrac{x-2}{x-1}\right)\cdot\dfrac{x+1}{x}\)
\(=\dfrac{\left(x+2\right)\left(x-1\right)-\left(x-2\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x+1}{x}\)
\(=\dfrac{x^2+x-2-\left(x^2-x-2\right)}{x-1}\cdot\dfrac{1}{x}=\dfrac{2x}{x\left(x-1\right)}=\dfrac{2}{x-1}\)
c: Để A nguyên thì \(2⋮x-1\)
=>\(x-1\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{2;0;3;-1\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{2;3\right\}\)
d: A=2B
=>\(\dfrac{2}{x-1}=2\cdot\dfrac{3}{x^2-1}\)
=>\(\dfrac{1}{x-1}=\dfrac{3}{\left(x-1\right)\left(x+1\right)}\)
=>x+1=3
=>x=2(nhận)
