Question

Answer 1

a:Khi x=2 thì \(B=\dfrac{-10}{2-4}=\dfrac{-10}{-2}=5\)

b: \(A=\dfrac{x^2+3x+2-5x-1-x-5}{\left(x+1\right)\left(x+5\right)}\)

\(=\dfrac{x^2-3x-4}{\left(x+1\right)\left(x+5\right)}=\dfrac{x-4}{x+5}\)

c: \(P=A\cdot B=\dfrac{x-4}{x+5}\cdot\dfrac{-10}{x-4}=\dfrac{-10}{x+5}\)

Để P nguyên thì \(x+5\in\left\{1;-1;2;-2;5;-5;10;-10\right\}\)

hay \(x\in\left\{-4;-6;-3;-7;0;-10;5;-15\right\}\)

Answer 2

làm full 2Gp