d) Để P>1 thì P-1>0
\(\Leftrightarrow\dfrac{-2}{x+2}-1>0\)
\(\Leftrightarrow\dfrac{-2-x-2}{x+2}>0\)
\(\Leftrightarrow\dfrac{-x-4}{x+2}>0\)
\(\Leftrightarrow\dfrac{x+4}{x+2}< 0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+4>0\\x+2< 0\end{matrix}\right.\Leftrightarrow-4< x< -2\)
e) Để P nguyên thì \(-2⋮x+2\)
\(\Leftrightarrow x+2\in\left\{1;-1;2;-2\right\}\)
\(\Leftrightarrow x\in\left\{-1;-3;0;-4\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{-1;-3;-4\right\}\)
a) ĐKXĐ: \(x\notin\left\{2;-2;0\right\}\)
Ta có: \(P=\dfrac{x^2-4x+4}{x^2-4}-\dfrac{x}{x^2+2x}\)
\(=\dfrac{\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x}{x\left(x+2\right)}\)
\(=\dfrac{x-2-x}{x+2}\)
\(=\dfrac{-2}{x+2}\)
b) Thay x=3 vào P, ta được:
\(P=\dfrac{-2}{3+2}=\dfrac{-2}{5}\)
c) Để P=2 thì x+2=-1
hay x=-3(nhận)
d)Ta có:\(P>1\Leftrightarrow\dfrac{-2}{x+2}>1\Leftrightarrow-2>x+2\Leftrightarrow x< -4\)
e)Để P nguyên thì x+2∈Ư(-2)={1;-1;2;-2}
⇔ x={-1;-3;0;-4}
Mà x=0 ko tm ĐKXĐ => x={-1;-3;-4}