a: \(A=\left(\frac{x^2}{x^3-4x}+\frac{6}{6-3x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(=\left(\frac{x^2}{x\left(x-2\right)\left(x+2\right)}-\frac{6}{3\left(x-2\right)}+\frac{1}{x+2}\right):\frac{\left(x-2\right)\left(x+2\right)+10-x^2}{x+2}\)
\(=\left(\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2}{x-2}+\frac{1}{x+2}\right):\frac{x^2-4+10-x^2}{x+2}\)
\(=\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{6}=\frac{2x-2-2x-4}{6\left(x-2\right)}=\frac{-1}{x-2}\)
b: |x-2|=4
=>x-2=4 hoặc x-2=-4
=>x=6(nhận) hoặc x=-2(loại)
Thay x=6 vào A, ta được:
\(A=\frac{-1}{6-2}=-\frac14\)
c: A=2
=>\(-\frac{1}{x-2}=2\)
=>\(x-2=-\frac12\)
=>\(x=-\frac12+2=\frac32\) (nhận)
d: A<0
=>\(\frac{-1}{x-2}<0\)
=>x-2>0
=>x>2
e: Để A nguyên thì -1⋮x-2
=>x-2∈{1;-1}
=>x∈{3;1}
