\(\sqrt{x-2\sqrt{x}+1}+\sqrt{x}=1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x}-1\right)^2}+\sqrt{x}=1\)
\(\Leftrightarrow|\sqrt{x}-1|+\sqrt{x}=1\)
TH1: \(|\sqrt{x}-1|=\sqrt{x}-1\) vs \(x\ge1\)
\(\Rightarrow\sqrt{x}-1=1\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\) (tm)
TH2: \(|\sqrt{x}-1|=1-\sqrt{x}\) vs \(x< 1\)
\(\Rightarrow1-\sqrt{x}=1\Leftrightarrow x=0\) (ktm)
Vậy x = 4
=.= hok tốt!!
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