Question

giúp e bai 1

Answer 1

a:

ĐKXĐ: \(x\notin\left\{0;5;-5;-1\right\}\)

 \(P=\left(\dfrac{15-x}{x^2-25}+\dfrac{2}{x+5}\right):\dfrac{x+1}{2x^2-10x}\)

\(=\dfrac{15-x+2\left(x-5\right)}{\left(x-5\right)\left(x+5\right)}\cdot\dfrac{2x\left(x-5\right)}{x+1}\)

\(=\dfrac{15-x+2x-10}{x+5}\cdot\dfrac{2x}{x+1}\)

\(=\dfrac{x+5}{x+5}\cdot\dfrac{2}{x+1}=\dfrac{2x}{x+1}\)

b: \(P=\dfrac{2x}{x+1}=\dfrac{2x+2-2}{x+1}=2-\dfrac{2}{x+1}\)

Để P nguyên thì \(-2⋮x+1\)

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

=>\(x\in\left\{0;-2;1;-3\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{-2;1;-3\right\}\)