Question

Answer 1

a: \(A=\dfrac{3\left(x+5\right)}{\left(x+5\right)^2}=\dfrac{3}{x+5}\)

|x+2|=3

=>x+2=3 hoặc x+2=-3

=>x=-5(loại) hoặc x=1(nhận)

Khi x=1 thì \(A=\dfrac{3}{1+5}=\dfrac{3}{6}=\dfrac{1}{2}\)

b: \(B=\dfrac{x}{x+3}+\dfrac{2x}{x-3}-\dfrac{3x^2+9}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x^2-3x+2x^2+6x-3x^2-9}{\left(x-3\right)\left(x+3\right)}=\dfrac{-3x-9}{\left(x+3\right)\left(x-3\right)}=\dfrac{-3}{x-3}\)

c: \(P=\dfrac{A}{B}=\dfrac{3}{x+5}:\dfrac{-3}{x-3}=\dfrac{-x+3}{x+5}\)

Để P nguyên thì \(-x-5+8⋮x+5\)

=>\(x+5\in\left\{1;-1;2;-2;4;-4;8;-8\right\}\)

=>\(x\in\left\{-4;-6;-7;-1;-9;-13\right\}\)