Lời giải:
a.
\(A=\left[\frac{x}{(x-2)(x+2)}+\frac{x-2}{(x+2)(x-2)}-\frac{2(x+2)}{(x-2)(x+2)}\right]: \frac{2}{x+2}\\ =\frac{x+x-2-2(x+2)}{(x+2)(x-2)}.\frac{x+2}{2}=\frac{-6}{(x+2)(x-2)}.\frac{x+2}{2}=\frac{3}{2-x}\)
b.
Khi $x=-4$ thì: $A=\frac{3}{2-(-4)}=\frac{3}{6}=\frac{1}{2}$
c.
Với $x$ nguyên, để $A$ nguyên thì $-3\vdots 2-x$
$\Rightarrow 2-x\in\left\{1; -1; 3; -3\right\}$
$\Rightarrow x\in\left\{1; 3; -1; 5\right\}$
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