Bài 1:
a:
ĐKXĐ: x<>2
\(P=\dfrac{2x^2-3x-2}{x^3-2x^2+2x-4}\)
\(=\dfrac{2x^2-4x+x-2}{x^2\left(x-2\right)+2\left(x-2\right)}\)
\(=\dfrac{\left(x-2\right)\left(2x+1\right)}{\left(x-2\right)\cdot\left(x^2+2\right)}=\dfrac{2x+1}{x^2+2}\)
b: |x-1|=4
=>\(\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\left(nhận\right)\\x=-3\left(nhận\right)\end{matrix}\right.\)
Thay x=-3 vào P, ta được:
\(P=\dfrac{2\cdot\left(-3\right)+1}{\left(-3\right)^2+2}=\dfrac{-5}{11}\)
Thay x=5 vào P, ta được:
\(P=\dfrac{2\cdot5+1}{5^2+2}=\dfrac{11}{27}\)