\(x^2-\left|x\right|=6\)
Nếu \(x\ge0\):
\(x^2-\left|x\right|=6\\ \Leftrightarrow x^2-x=6\\ \Leftrightarrow x\cdot\left(x-1\right)=6\\ \Rightarrow\left[{}\begin{matrix}x\cdot\left(x-1\right)=3\cdot2\\x\cdot\left(x-1\right)=\left(-2\right)\cdot\left(-3\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\left(t.m\right)\\x=-2\left(k.t.m\right)\end{matrix}\right.\)
Nếu \(x< 0\)
\(x^2-\left|x\right|=6\\ \Leftrightarrow x^2-\left(-x\right)=6\\ \Leftrightarrow x^2+x=6\\ \Leftrightarrow x\cdot\left(x+1\right)=6\\ \Rightarrow\left[{}\begin{matrix}x\cdot\left(x+1\right)=2\cdot3\\x\cdot\left(x+1\right)=\left(-3\right)\cdot\left(-2\right)\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\left(k.t.m\right)\\x=-3\left(t.m\right)\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=-3\)
\(x^2-\left|x\right|=6\\ \Rightarrow\left[{}\begin{matrix}x^2-x=6\left(với\: x\ge0\right)\\x^2+x=6\left(với\: x< 0\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}\left(x+2\right)\left(x-3\right)=0\\\left(x+2\right)\left(x+3\right)=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x+2=0\\x-3=0\\x+3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-2\left(loại\right)\\x=3\left(nhận\right)\\x=-3\left(nhận\right)\end{matrix}\right.\)
vậy....
