\(x^2-x=6\\ \Leftrightarrow x^2-x-6=0\left(1\right)\)
Ta có:
\(x^2-x-6=\left(x^2-2.x.\frac{1}{2}+\frac{1}{4}\right)+\frac{23}{4}\\ =\left(x-\frac{1}{2}\right)^2+\frac{23}{4}\ge\frac{23}{4}>0\forall x\)
\(\Rightarrow\) Phương trình (1) không xảy ra
Vậy phương trình \(x^2-x=6\) vô nghiệm
\(x^2-x=6\)
\(\Leftrightarrow x^2-x-6=0\)
\(\Leftrightarrow x^2-3x+2x-6=0\)
\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
