ĐKXĐ : \(\left\{{}\begin{matrix}\sqrt{2x+1}-1\ne0\\2x+1\ge0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}2x+1\ne1\\2x+1\ge0\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x\ne0\\x\ge-\frac{1}{2}\end{matrix}\right.\)
=> \(x\ge-\frac{1}{2}\)
Ta có : \(\frac{x}{\sqrt{2x+1}-1}=1-x\)
=> \(x=\left(\sqrt{2x+1}-1\right)\left(1-x\right)\)
=> \(x=\sqrt{2x+1}-1-x\sqrt{2x+1}+x\)
=> \(\sqrt{2x+1}-1=x\sqrt{2x+1}\)
=> \(\left(x-1\right)\sqrt{2x+1}=-1\)
=> \(\sqrt{2x+1}=-\frac{1}{x-1}\)
<=> \(x< 1\)
<=> \(-\frac{1}{2}\le x< 1\)
=> \(2x+1=\frac{1}{x^2-2x+1}\)
=> \(2x^3-3x^2+1=1\)
=> \(x^2\left(2x-3\right)=0\)
=> \(\left[{}\begin{matrix}x=0\\x=\frac{3}{2}\end{matrix}\right.\) ( KTM )
Vậy phương trình vô nghiệm .
Lời giải:
ĐK: $x\neq 0; x\geq \frac{-1}{2}$
PT $\Rightarrow x=(1-x)(\sqrt{2x+1}-1)$
$\Leftrightarrow x=(1-x).\frac{2x+1-1}{\sqrt{2x+1}+1}$
$\Leftrightarrow x(\sqrt{2x+1}+1)=2x(1-x)$
$\Rightarrow \sqrt{2x+1}+1=2(1-x)$ (do $x\neq 0$)
$\Leftrightarrow \sqrt{2x+1}=1-2x$
\(\Rightarrow \left\{\begin{matrix} 1-2x\geq 0\\ 2x+1=(1-2x)^2=4x^2-4x+1\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\leq \frac{1}{2}\\ 4x^2-6x=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\leq \frac{1}{2}\\ x(2x-3)=0\end{matrix}\right.\)
Với điều kiện $x\leq \frac{1}{2}; x\neq 0$ thì ta thấy PT vô nghiệm.
