Question

Giải giúp em với ạ

Answer 1

a: \(P=\left(\dfrac{x}{x-3}-\dfrac{1}{x+3}+\dfrac{x^2-1}{9-x^2}\right):\dfrac{2}{x+3}\)

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

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

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

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

b: |x+4|=1

=>\(\left[{}\begin{matrix}x+4=1\\x+4=-1\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=-3\left(loại\right)\\x=-5\left(nhận\right)\end{matrix}\right.\)

Thay x=-5 vào P, ta được:

\(P=\dfrac{-5+2}{-5-3}=\dfrac{-3}{-8}=\dfrac{3}{8}\)

c: Để P là số nguyên thì \(x+2⋮x-3\)

=>\(x-3+5⋮x-3\)

=>\(5⋮x-3\)

=>\(x-3\in\left\{1;-1;5;-5\right\}\)

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