Đáp án:
a) $A=\dfrac{2}{x-3}$
b) $x=7$
Giải thích các bước giải:
ĐKXĐ: $x\ne ±3$
a) $A=\dfrac{3x+15}{x^2-9}+\dfrac{1}{x+3}-\dfrac{2}{x-3}$
$=\dfrac{3x+15}{x^2-9}+\dfrac{x-3}{x^2-9}-\dfrac{2(x+3)}{x^2-9}$
$=\dfrac{3x+15+x-3-2x-6}{x^2-9}$
$=\dfrac{2x+6}{x^2-9}$
$=\dfrac{2(x+3)}{(x-3)(x+3)}$
$=\dfrac{2}{x-3}$
b) $A=\dfrac{1}{2}$
$⇒\dfrac{2}{x-3}=\dfrac{1}{2}$
$⇒x-3=4$
$⇒x=7$
Vậy $A=\dfrac{1}{2}$ khi $x=7$.
