Bài 3:
a: \(B=\dfrac{x}{x-1}+\dfrac{3}{x+1}-\dfrac{6x-4}{x^2-1}\)
\(=\dfrac{x}{x-1}+\dfrac{3}{x+1}-\dfrac{6x-4}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)-6x+4}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+x+3x-3-6x+4}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
b: x-1=-3
=>x=-2(nhận)
Thay x=-2 vào A, ta được:
\(A=\dfrac{2\cdot\left(-2\right)-3}{-2+1}=\dfrac{-4-3}{-1}=7\)
c: P=A:B
\(=\dfrac{2x-3}{x+1}:\dfrac{x-1}{x+1}\)
\(=\dfrac{2x-3}{x+1}\cdot\dfrac{x+1}{x-1}=\dfrac{2x-3}{x-1}\)
Để \(P=\dfrac{5}{2}\) thì \(\dfrac{2x-3}{x-1}=\dfrac{5}{2}\)
=>5(x-1)=2(2x-3)
=>5x-5=4x-6
=>5x-4x=-6+5
=>x=-1(loại)
d: Để P là số nguyên thì \(2x-3⋮x-1\)
=>\(2x-2-1⋮x-1\)
=>\(-1⋮x-1\)
=>\(x-1\in\left\{1;-1\right\}\)
=>\(x\in\left\{2;0\right\}\)
