a: \(A=\dfrac{x}{x-2}-\dfrac{2x+1}{x+2}-\dfrac{x^2+8}{4-x^2}\)
\(=\dfrac{x}{x-2}-\dfrac{2x+1}{x+2}+\dfrac{x^2+8}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x\left(x+2\right)-\left(2x+1\right)\left(x-2\right)+x^2+8}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{x^2+2x-\left(2x^2-4x+x-2\right)+x^2+8}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{2x^2+2x+8-2x^2+3x+2}{\left(x-2\right)\left(x+2\right)}=\dfrac{5x+10}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{5\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{5}{x-2}\)
b: Khi x=-6 thì \(A=\dfrac{5}{-6-2}=\dfrac{5}{-8}=-\dfrac{5}{8}\)
c: Để A=2 thì \(\dfrac{5}{x-2}=2\)
=>\(x-2=\dfrac{5}{2}\)
=>\(x=\dfrac{9}{2}\left(nhận\right)\)
d: \(M=A:B=\dfrac{5}{x-2}:\dfrac{5}{x+1}=\dfrac{x+1}{x-2}\)
Để M là số nguyên thì \(x+1⋮x-2\)
=>\(x-2+3⋮x-2\)
=>\(3⋮x-2\)
=>\(x-2\in\left\{1;-1;3;-3\right\}\)
=>\(x\in\left\{3;1;5;-1\right\}\)
Kết hợp ĐKXĐ, ta được:
\(x\in\left\{3;1;5\right\}\)
