a) Với đkxđ: $x\neq2$ , $x\neq-3$ ta có:
P= $\frac{9-x^{2}}{x^{2}+x+6}-\frac{x-3}{2-x}-\frac{x-2}{x+3}$
P= $\frac{(3-x)(3+x)}{x^{2}-2x+3x-6}-\frac{x-3}{2-x}-\frac{x-2}{x+3}$
P= $\frac{(3-x)(3+x)}{x(x-2)+3(x-2)}-\frac{x-3}{2-x}-\frac{x-2}{x+3}$
P= $\frac{(3-x)(3+x)}{(x-2)(x+3)}-\frac{(x-3)(x+3)}{(2-x)(x+3)}-\frac{(x-2)(x-2)}{(x+3)(x-2)}$
P= $\frac{9+3x-3x-2x+2x+3x-3x-9-x^{2}+2x+2x-4}{(x-2)(x+3)}$
P= $\frac{-x^{2}+4x-4}{(x-2)(x+3)}$
P= $\frac{-x^{2}+2x+2x-4}{(x-2)(x+3)}$
P= $\frac{(x-2)(x+2)}{(x-2)(x+3)}$
P= $\frac{x+2}{x+3}$
b) Với đkxđ: $x\neq2$ , $x\neq-3$ ta có:
$\frac{x+2}{x+3}=\frac{-3}{4}$
$⇔4(x+2)=-3(x+3)$
$⇔ 4x+8=-3x-9$
$⇔ 7x=-17$
$⇔ x=\frac{-17}{7}(tmđk)$
c) Với đkxđ: $x\neq2$ , $x\neq-3$ ta có:$\frac{x+2}{x+3}>-2$
$⇔\frac{x+2}{x+3}+2>0$
$⇔\frac{x+2}{x+3}+\frac{2(x+3)}{x+3}$
$⇔\frac{x+2+2x+6}{x+3}$
$⇔\frac{3x+8}{x+3}$
$TH1: \left \{ {{3x+8>0} \atop {x+3>0}} \right.⇔\left \{ {x>{\frac{-8}{3}} \atop {x>-3}} \right.⇔ x>\frac{-8}{3}(tmđk)$
$TH2: \left \{ {{3x+8<0} \atop {x+3<0}} \right.⇔\left \{ {x<{\frac{-8}{3}} \atop {x<-3}} \right.⇔ x<-3(tmđk)$
Vậy $x>\frac{-8}{3}hoặc x<-3$
d) $\frac{x+2}{x+3}=\frac{x+3-3+2}{x+3}=1-\frac{5}{x+3}$
x ∈ Z ⇔ P ∈ Z ⇔ x+3 ∈ Ư {5}
x+3 ∈ {5;-5;1;-1}
ta có bảng: (dưới)
Vậy để P ∈ Z thì x ∈ {-2; -4; -8}
làm đúng 2 tiếng luôn hic :<
