\(x^2+2x-3=\left|2x+1\right|\)
\(\Rightarrow\left\{{}\begin{matrix}2x+1=x^2+2x-3\\2x+1=-x^2-2x+3\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x^2=4\\x^2+4x-2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\pm2\\x^2+4x+4-6=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\pm2\\\left(x+2\right)^2-\left(\sqrt{6}\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\pm2\\\left(x+2+\sqrt{6}\right)\left(x+2-\sqrt{6}\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\pm2\\\left[{}\begin{matrix}x=-\sqrt{6}-2\\x=\sqrt{6}-2\end{matrix}\right.\end{matrix}\right.\)
Vậy.......................
Vì \(x\in R\) nên \(x< 0\) hoặc \(x>0\)
Nếu \(x>0\) thì ta có :
Vì \(x>0\) nên \(\left|2x+1\right|=2x+1\)
\(x^2+2x-3=2x+1\)
\(x^2+2x=2x+4\)
Suy ra :\(x^2=4\Rightarrow x=2\) (thỏa mãn)
Nếu \(x< 0\) thì ta có :
Vì \(x< 0\) nên \(\left|2x+1\right|=2\left(-x\right)-1\)
\(x^2+2x-3=2\left(-x\right)-1\)
\(x^2+2x=2\left(-x\right)+2\)
\(x^2\times\left(-1\right)+2x=2x+2\)
Suy ra : \(x^2\times\left(-1\right)=2\Rightarrow x^2=-2\)(không thỏa mãn)
Vậy số cần tìm là 2
