Đáp án:
a.$\dfrac{x^2}{x^2+2x}-\dfrac{2}{x}+\dfrac{2}{x+2}$$=\dfrac{x-2}{x}$
b.$\dfrac{12-3x}{x^2+4x+4}:\dfrac{4x-16}{x+2}$$=\dfrac{-3}{4(x+2)}$
Giải thích các bước giải:
a.$\dfrac{x^2}{x^2+2x}-\dfrac{2}{x}+\dfrac{2}{x+2}$
$=\dfrac{x^2}{x(x+2}-\dfrac{2}{x}+\dfrac{2}{x+2}$
$=\dfrac{x}{x+2}+\dfrac{2}{x+2}-\dfrac{2}{x}$
$=\dfrac{x+2}{x+2}-\dfrac{2}{x}$
$=1-\dfrac{2}{x}$
$=\dfrac{x-2}{x}$
b.$\dfrac{12-3x}{x^2+4x+4}:\dfrac{4x-16}{x+2}$
$=\dfrac{3(4-x)}{(x+2)^2}: \dfrac{4(x-4)}{x+2}$
$=\dfrac{-3(x-4)}{(x+2)^2}.\dfrac{x+2}{4(x-4)}$
$=\dfrac{-3}{4(x+2)}$
