a: \(P=\dfrac{1}{x-2}-\dfrac{x^2+8}{x^3-8}-\dfrac{4}{x^2+2x+4}\)
\(=\dfrac{1}{x-2}-\dfrac{x^2+8}{\left(x-2\right)\left(x^2+2x+4\right)}-\dfrac{4}{x^2+2x+4}\)
\(=\dfrac{x^2+2x+4-x^2-8-4x+8}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{-2x+4}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{-2}{x^2+2x+4}\)
b: \(2x^2+x-6=0\)
=>\(2x^2+4x-3x-6=0\)
=>(x+2)(2x-3)=0
=>\(\left[{}\begin{matrix}x=-2\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)
Thay x=-2 vào P, ta được:
\(P=\dfrac{-2}{\left(-2\right)^2+2\cdot\left(-2\right)+4}=\dfrac{-2}{4}=-\dfrac{1}{2}\)
Thay x=3/2 vào P, ta được:
\(P=\dfrac{-2}{\left(\dfrac{3}{2}\right)^2+2\cdot\dfrac{3}{2}+4}=\dfrac{-2}{\dfrac{9}{4}+7}=-2:\dfrac{37}{4}=-\dfrac{8}{37}\)
c: \(x^2+2x+4=x^2+2x+1+3=\left(x+1\right)^2+3>=3\)
=>\(P=\dfrac{-2}{x^2+2x+4}< =\dfrac{-2}{3}< 0\forall x\) thỏa mãn ĐKXĐ
d: \(P< =-\dfrac{2}{3}\forall x\) thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x+1=0
=>x=-1
