a) \(ĐKXĐ:x\ne\pm4;x\ne-2\)
\(P=\left(\frac{8}{x^2-16}+\frac{1}{x+4}\right):\frac{1}{x^2-2x-8}\)
\(\Leftrightarrow P=\left(\frac{8}{\left(x-4\right)\left(x+4\right)}+\frac{1}{x+4}\right):\frac{1}{\left(x-4\right)\left(x+2\right)}\)
\(\Leftrightarrow P=\frac{8+x-4}{\left(x-4\right)\left(x+4\right)}:\frac{1}{\left(x-4\right)\left(x+2\right)}\)
\(\Leftrightarrow P=\frac{x+4}{\left(x-4\right)\left(x+4\right)}:\frac{1}{\left(x-4\right)\left(x+2\right)}\)
\(\Leftrightarrow P=\frac{1}{x-4}.\left(x-4\right)\left(x+2\right)\)
\(\Leftrightarrow P=\frac{\left(x-4\right)\left(x+2\right)}{\left(x-4\right)}\)
\(P=x+2\)
b) Ta có :
\(x^2-9x+20=0\)
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}P=x+2=5+2=7\\P=x+2=4+2=6\end{cases}}\)
Vậy \(P\in\left\{7;6\right\}\)
