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Điều kiện $x \ge 1/2.$
PT
$\Leftrightarrow x^{2}= 2x+2\sqrt{2x-1}$
$\Leftrightarrow x^{2}=( 2x-1)+2\sqrt{2x-1}+1$
$\Leftrightarrow x^{2}=\left ( \sqrt{2x-1}+1 \right )^2$
Do $x \ge 1/2$ nên $x>0, \sqrt{2x-1}+1>0$ ta có
$\Leftrightarrow x= \sqrt{2x-1}+1$
$\Leftrightarrow x-1= \sqrt{2x-1}$
$\Leftrightarrow \begin{cases}(x-1)^2=2x-1 \\ x \ge 1 \end{cases}$
$\Leftrightarrow \begin{cases}x^2-4x+2=0 \\ x \ge 1 \end{cases}$
$\Leftrightarrow \boxed{x=2+\sqrt 2} $
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