Bài 4:
a: Thay x=-1 vào A, ta được:
\(A=\dfrac{-1+1}{-1-2}=\dfrac{0}{-3}=0\)
b: \(B=\dfrac{3x-4}{x^2-2x}-\dfrac{x+2}{x}-\dfrac{x-1}{x-2}\)
\(=\dfrac{3x-4}{x\left(x-2\right)}-\dfrac{x+2}{x}-\dfrac{x-1}{x-2}\)
\(=\dfrac{3x-4-\left(x+2\right)\left(x-2\right)-x\left(x-1\right)}{x\left(x-2\right)}\)
\(=\dfrac{3x-4-x^2+4-x^2+x}{x\left(x-2\right)}\)
\(=\dfrac{-2x^2+4x}{x\left(x-2\right)}=\dfrac{-2x\left(x-2\right)}{x\left(x-2\right)}=-2\)
c: Để A là số nguyên thì \(x+1⋮x-2\)
=>\(x-2+3⋮x-2\)
=>\(3⋮x-2\)
=>\(x-2\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{3;1;5;-1\right\}\)
Bài 5:
a: ĐKXĐ: \(x\notin\left\{0;2;-2;-1\right\}\)
\(A=\left(\dfrac{4x}{x^2+2x}-\dfrac{2}{2-x}+\dfrac{6-5x}{x^2-4}\right):\dfrac{x+1}{x-2}\)
\(=\left(\dfrac{4x}{x\left(x+2\right)}+\dfrac{2}{x-2}+\dfrac{6-5x}{\left(x+2\right)\left(x-2\right)}\right)\cdot\dfrac{x-2}{x+1}\)
\(=\left(\dfrac{4}{x+2}+\dfrac{2}{x-2}+\dfrac{6-5x}{\left(x+2\right)\left(x-2\right)}\right)\cdot\dfrac{x-2}{x+1}\)
\(=\dfrac{4\left(x-2\right)+2\left(x+2\right)+6-5x}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x-2}{x+1}\)
\(=\dfrac{4x-8+2x+4+6-5x}{\left(x+2\right)}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{x+2}{\left(x+2\right)\left(x+1\right)}=\dfrac{1}{x+1}\)
b: Để A là số nguyên thì \(1⋮x+1\)
=>\(x+1\in\left\{1;-1\right\}\)
=>\(x\in\left\{0;-2\right\}\)
Kết hợp ĐKXĐ, ta được; \(x\in\varnothing\)
