a: \(P=\left(\dfrac{x-2}{x+2}+\dfrac{x}{x-2}+\dfrac{2x+4}{4-x^2}\right)\cdot\left(1+\dfrac{5}{x-3}\right)\)
\(=\left(\dfrac{x-2}{x+2}+\dfrac{x}{x-2}-\dfrac{2x+4}{\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{x-3+5}{x-3}\)
\(=\dfrac{\left(x-2\right)^2+x\left(x+2\right)-2x-4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+2}{x-3}\)
\(=\dfrac{x^2-4x+4+x^2+2x-2x-4}{\left(x-2\right)}\cdot\dfrac{1}{x-3}\)
\(=\dfrac{2x^2-4x}{\left(x-2\right)\cdot\left(x-3\right)}=\dfrac{2x}{x-3}\)
b: Thay x=-1 vào P, ta được:
\(P=\dfrac{2\cdot\left(-1\right)}{-1-3}=\dfrac{-2}{-4}=\dfrac{1}{2}\)
c: \(P=\dfrac{2}{3}\)
=>\(\dfrac{2x}{x-3}=\dfrac{2}{3}\)
=>\(\dfrac{x}{x-3}=\dfrac{1}{3}\)
=>3x=x-3
=>2x=-3
=>\(x=-\dfrac{3}{2}\left(nhận\right)\)
d: Để P là số tự nhiên thì \(\left\{{}\begin{matrix}\dfrac{2x}{x-3}>=0\\2x⋮x-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{x}{x-3}>=0\\2x-6+6⋮x-3\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>3\\x< =0\end{matrix}\right.\\x-3\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>3\\x< =0\end{matrix}\right.\\x\in\left\{4;2;5;1;6;0;9;-3\right\}\end{matrix}\right.\)
=>\(x\in\left\{4;5;6;0;9;-3\right\}\)
