x3-8+(x-2)(x+1)=0
(x-2)(x2+2x+4)+(x-2)(x+1)=0
(x-2)(x2+2x+4+x+1)=0
(x-2)(x2+3x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x^2+3x+5=0\end{matrix}\right.\)
+) x-2=0 => x=2
+) x2+3x+5=0
\(\left(x^2+2.x.\dfrac{3}{2}+\dfrac{9}{4}\right)+\dfrac{11}{4}=0\)
\(\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{4}=0\) mà \(\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{4}\ge\dfrac{11}{4}\forall x\)
=> phương trình vô nghiệm
Vậy phương trình có nghiệm x=2
`x^3-8+(x-2)(x+1)=0`
`=> x^3-2^3+(x-2)(x+1)=0`
`=> (x-2)(x^2+2x+4)+(x-2)(x+1)=0`
`=> (x-2)(x^2+2x+4+x+1)=0`
`=> (x-2)(x^2+3x+5)=0`
`=>` \(\left[{}\begin{matrix}x-2=0\\x^2+3x+5=0\end{matrix}\right.\)
`=>` \(\left[{}\begin{matrix}x=2\\\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{4}=0\end{matrix}\right.\)
`=>` \(\left[{}\begin{matrix}x=2\\\left(x+\dfrac{3}{2}\right)^2=-\dfrac{11}{4}\end{matrix}\right.\)
Có: \(\left(x+\dfrac{3}{2}\right)^2\ge0\forall x\)
`=> x=2`
