giup mik voi a lam het nhaa

Đáp án:

/ảnh/

Đáp án+Giải thích các bước giải:

$Bài 1:$

$a) (x-3)^2 -(3-x)(x+3)$
$ = (x-3)^2 +(x-3)(x+3)$

$ = (x-3)(x-3+x+3)$
$ = (x-3).2x$
$ b) 5x^2 -10xy +5y^2 -20z^2$
$ = 5(x^2 -2xy +y^2 -4z^2)$
$ = 5[(x^2-2xy +y^2) -4z^2]$
$ = 5[(x-y)^2 -(2z)^2]$
$ = 5(x-y-2z)(x-y+2z)$
$ Bài 2:$
$a) (2x-3y)^2$
$ = 4x^2 -12xy +9y^2$

$b) \frac{x+3}{x^2-1} -\frac{1}{x^2+x}$
$ = \frac{x+3}{(x-1)(x+1)} - \frac{1}{x(x+1)}$
$ = \frac{x(x+3)}{x(x-1)(x+1)} - \frac{x-1}{x(x+1)(x-1)}$
$ = \frac{x^2 +3x -x+1}{x(x-1)(x+1)}$
$ = \frac{x^2 +2x +1}{x(x-1)(x+1)}$
$ = \frac{(x+1)^2}{x(x-1)(x+1)}$
$ = \frac{x+1}{x(x-1)}$

$c) \frac{3x+9}{x^2-4} : \frac{x+3}{x-2}$
$ = \frac{3(x+3)}{(x-2)(x+2)} .\frac{x-2}{x+3}$
$ = \frac{3(x+3).(x-2)}{(x-2)(x+2).(x+3)}$
$ = \frac{3}{x+2}$
$#sotuyet$