Đáp án:
Giải thích các bước giải:
a) A=\((\frac{1}{x-2}-\frac{2x}{4-x^{2}}+\frac{1}{2+x})\cdot (\frac{2}{x}-1)\)
=\((\frac{1}{x-2}+\frac{2x}{x^{2}-4}+\frac{1}{x+2})\cdot (\frac{2-x}{x})\)
=\((\frac{x+2+2x+x-2}{(x-2)(x+2)})\cdot \frac{2-x}{x}\)
=\(\frac{4x}{(x-2)(x+2)}\cdot \frac{-(x+2)}{x}\)
=\(\frac{-4}{x-2}\)
b)A=\(\frac{-4}{x-2}=\frac{-4}{4-2}\)=-2
c) A=\(\frac{1}{2}⇒\frac{-4}{x-2}=\frac{1}{2}\)⇒ x-2=-8⇒ x=-6
d) Để \(\frac{-4}{x-2}∈Z⇒ -4\vdots x-2\)⇒ ƯC={ ±1,±2,±4}
· x-2=1⇒ x=3
·x-2=-1⇒ x=1
·x-2=2⇒x=4
x-2=-2⇒ x=0
x-2=4⇒ x=6
x-2=-4⇒ x=-2(loại)
Vậy S={ 0,1,3,4,6}
