Cho biểu thức: A=(1/(x-2)+2x/(x^2-4)+1/x+2)(2/x-1). a) Rút gọn biểu thức A. b) Tìm x để A = 1.

a) ĐKXĐ: $\{\begin{array}{l}x\ne 0\\ x-2\ne 0\\ x+2\ne 0\end{array}\iff \{\begin{array}{l}x\ne 0\\ x\ne \pm 2\end{array}$

Ta có: $A=(\frac{1}{x-2}+\frac{2x}{x^2-4}+\frac{1}{x+2})(\frac{2}{x}-1)$

$=[\frac{1}{x-2}+\frac{2x}{(x-2)(x+2)}+\frac{1}{x+2}].\frac{2-x}{x}$

$=[\frac{x+2}{(x-2)(x+2)}+\frac{2x}{(x-2)(x+2)}+\frac{x-2}{(x-2)(x+2)}].\frac{-(x-2)}{x}$

$=\frac{x+2+2x+x-2}{(x-2)(x+2)}.\frac{-(x-2)}{x}$

$=\frac{4x}{x+2}.\frac{-1}{x}=\frac{-4}{x+2}$.

Vậy $A=(\frac{1}{x-2}+\frac{2x}{x^2-4}+\frac{1}{x+2})(\frac{2}{x}-1)=\frac{-4}{x+2}$.

 b) Ta có A = 1 hay $\frac{-4}{x+2}=1$.

Û x + 2 = − 4

Û x = − 6 (TMĐK).

Vậy để A =1 thì x = − 6.