a) Đkxđ:
$x^2-9\neq 0 ⇔ x\neq±3$
$6x-12\neq 0 ⇔ x\neq 2$
Rút gọn:
B= ($\frac{(x+3)(x+3)}{x^2-9} +\frac{-2x^2+6}{x^2-9}+ \frac{x(x-3)}{x^2-9}$ ). $\frac{x^2-9}{3(x-2)}$
=$\frac{(x+3)^2-2x^2+6+x(x-3)}{x^2-9}$ .$\frac{x^2-9}{3(x-2)}$
=$\frac{3x+15}{3(x-2)}$ =$\frac{x+5}{x-2}$
b) Ta có: |x+1|=2⇔\(\left[ \begin{array}{l}x+1=2\\x+1=-2\end{array} \right.\) ⇔\(\left[ \begin{array}{l}x=1\\x=-3\end{array} \right.\)
Với x=1: B=-6
Với x=-3: B= $\frac{-2}{5}$
c) Ta có: B= $\frac{x+5}{x-2}=1+\frac{7}{x-2}$
Để B nguyên thì $\frac{7}{x-2}$ phải nguyên
⇔ (x-2) ∈U(7)={±1;±7}
Nếu x-2= -1 ⇒x=1
Nếu x-2= 1 ⇒x=3
Nếu x-2= -7 ⇒x=-5
Nếu x-2= 7 ⇒x=9
Vậy x∈{1;3;-5;9}