Giải thích các bước giải:
Bài 9:
a.Ta có:
$\dfrac{6}{x^2+4x}+\dfrac{3}{2x+8}$
$=\dfrac{6}{x(x+4)}+\dfrac{3}{2(x+4)}$
$=\dfrac{12}{2x(x+4)}+\dfrac{3x}{2x(x+4)}$
$=\dfrac{12+3x}{2x(x+4)}$
$=\dfrac{3(4+x)}{2x(4+x)}$
$=\dfrac3{2x}$
b.Ta có:
$\dfrac{x+15}{x^2-9}+\dfrac{2}{x+3}$
$=\dfrac{x+15}{(x-3)(x+3)}+\dfrac{2(x-3)}{(x-3)(x+3)}$
$=\dfrac{x+15+2(x-3)}{(x-3)(x+3)}$
$=\dfrac{3x+9}{(x-3)(x+3)}$
$=\dfrac{3(x+3)}{(x-3)(x+3)}$
$=\dfrac3{x-3}$
c.Ta có:
$3x(x^3-2x)$
$=3x^4-6x^2$
d.Ta có:
$(x^3+x^2-2x):(x+2)$
$=(x^3+2x^2-x^2-2x):(x+2)$
$=(x^2(x+2)-x(x+2)):(x+2)$
$=(x^2-x)(x+2):(x+2)$
$=x^2-x$
