Question

Answer 1

Bài 2:

a: Khi x=2 thì \(A=\dfrac{2^2-9}{3\left(2+5\right)}=\dfrac{-5}{21}\)

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

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

Để P là số nguyên thì \(-x-5+2⋮x+5\)

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

hay \(x\in\left\{-4;-6;-7\right\}\)