Question

giải phương trình : \(|x^2-1|=2x+1\)

Answer 1

\(\circledast-1\le x^2-1< 0\Leftrightarrow0\le x^2\le1\)

\(\Rightarrow\left|x^2-1\right|=-x^2+1=2x+1\)

\(\Rightarrow-x^2-2x=0\Leftrightarrow-x\left(x+2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\) (loại)

\(\circledast x^2-1\ge0\Leftrightarrow x^2\ge1\)

\(\Rightarrow\left|x^2-1\right|=x^2-1=2x+1\)

\(\Rightarrow x^2+2x=2\)

\(\Rightarrow x^2+2x+1=\left(x+1\right)^2=3\)

\(\Rightarrow\left[{}\begin{matrix}x+1=\sqrt{3}\\x+1=-\sqrt{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{3}-1\left(loai\right)\\x=-\sqrt{3}-1\left(thoa\right)\end{matrix}\right.\)

Vậy........................