Đáp án: a) $ĐKXĐ : x \neq ±2, x \neq -\dfrac{6}{13}$ , $B = \dfrac{2}{x+2}$
b) $B<0$ khi $x<-2$ $\text{và}$ $B>0$ khi $x >-2,x \neq -\dfrac{6}{13},x \neq 2$
Giải thích các bước giải:
a) $ĐKXĐ : x \neq ±2, x \neq \dfrac{-6}{13}$
$\text{Rút gọn B:}$
$B = \bigg(\dfrac{1+2x}{4+2x}-\dfrac{x}{3x-6}+\dfrac{2x^2}{12-3x^2}\bigg) . \dfrac{24-12x}{6+13x}$
$ = \bigg[\dfrac{(1+2x).(x-2).3}{6.(x+2).(x-2)} - \dfrac{2x.(x+2)}{6.(x-2).(x+2)}-\dfrac{4x^2}{6.(x-2).(x+2)}\bigg] . \dfrac{12.(2-x)}{6+13x}$
$ = \bigg[\dfrac{6x^2-9x-6-2x^2-4x-4x^2}{6.(x-2).(x+2)}\bigg]. \dfrac{12.(2-x)}{6+13x}$
$ = \dfrac{-(6+13x)}{6.(x-2).(x+2)}.\dfrac{12.(2-x)}{6+13x}$
$ = \dfrac{6+13x}{6.(2-x).(x+2)}.\dfrac{12.(2-x)}{6+13x}$
$ = \dfrac{2}{x+2}$
Vậy $B = \dfrac{2}{x+2}$ $\text{với}$ $x \neq ±2,x \neq -\dfrac{6}{13}$
b) $\text{Theo câu a) ta có :}$ $B = \dfrac{2}{x+2}$ $\text{với}$ $x \neq ±2,x \neq -\dfrac{6}{13}$
$\text{Để B>0 thì }$ $\dfrac{2}{x+2} > 0 $ $\text{mà 2>0 nên suy ra}$
$x+2 > 0 $ $⇔ x > -2$ $\text{kết hợp với ĐKXĐ}$
$⇒ x>-2, x \neq -\dfrac{6}{13}, x \neq 2$
$\text{Ngược lại, để}$ $B < 0 ⇔ x <-2$
$\text{Vậy}$ $B<0 ⇔ x<-2$ $\text{và}$ $B>0⇔ x >-2,x \neq -\dfrac{6}{13},x \neq 2$
