a) \(P=\left(\frac{x-4}{x^3-1}+\frac{1}{x-1}\right):\left(1-\frac{x-8}{x^2+x+1}\right)\)
\(P=\left(\frac{x-4}{\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}\right):\left(\frac{x^2+x+1}{x^2+x+1}-\frac{x-8}{x^2+x+1}\right)\)
\(P=\frac{x-4+x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}:\frac{x^2+x+1-x+8}{x^2+x+1}\)
\(P=\frac{x^2+2x-3}{\left(x-1\right)\left(x^2+x+1\right)}\cdot\frac{x^2+x+1}{x^2+9}\)
\(P=\frac{\left(x+3\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+9\right)}\)
\(P=\frac{x+3}{x^2+9}\)
b) \(x^2-3x+2=0\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(loai\right)\\x=2\left(chon\right)\end{matrix}\right.\)
Thay \(x=2\)ta có : \(P=\frac{2+3}{2^2+9}=\frac{5}{13}\)
