+) Nếu \(8-x\ge0\Rightarrow x\le8\)
\(\Rightarrow8-x=x^2+x\)
\(\Leftrightarrow8-x-x^2-x=0\)
\(\Leftrightarrow8-2x-x^2=0\)
\(\Leftrightarrow-\left(x^2+2x-8\right)=0\)
\(\Leftrightarrow-\left(x^2-2x+4x-8\right)=0\)
\(\Leftrightarrow-\left[x\left(x-2\right)+4\left(x-2\right)\right]=0\)
\(\Leftrightarrow-\left(x-2\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\) ( thoả mãn \(x\le8\) )
+) Nếu \(8-x< 0\Rightarrow x>8\)
\(\Rightarrow x-8=x^2+x\)
\(\Leftrightarrow-x^2-8=0\)
\(\Leftrightarrow-\left(x^2+8\right)=0\)
Ta thấy \(x^2+8>0\forall x\)
Nên \(-\left(x^2+8\right)< 0\forall x\)
Vậy phương trình này vô nghiệm.
Do đó x = 2 hoặc x = -4
