\(\sqrt{x+2\sqrt{x-1}}=x-1\)
ĐK:\(x\ge 1\)
\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}=x-1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}=x-1\)
\(\Leftrightarrow\sqrt{x-1}+1=x-1\)
\(\Leftrightarrow\sqrt{x-1}=x-2\)
\(\Leftrightarrow x-1=x^2-4x+4\)
\(\Leftrightarrow-x^2+5x-5=0\Leftrightarrow x=\dfrac{\sqrt{5}}{2}+\dfrac{5}{2}\)
\(\sqrt{x+2\sqrt{x-1}}=x-1\)
ĐK XĐ
(đk1) \(x-1\ge0\Rightarrow x\ge1\)
\(\Leftrightarrow\sqrt{\left(x-1\right)+2\sqrt{x-1}+1}=x-1\)
\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}=x-1\)
\(\Leftrightarrow\left|\sqrt{x-1}+1\right|=x-1\)
Có \(\sqrt{x-1}+1>0\forall x\ge1\)
\(\Leftrightarrow\sqrt{x-1}+1=x-1\)
\(\Leftrightarrow\sqrt{x-1}=x-2\)
đk của nghiệm \(x\ge2\)
\(\Leftrightarrow x-1=x^2-4x+4\)
\(\Leftrightarrow x^2-5x+5=0\)
\(\Delta=25-4.5=5\)
\(x_1=\dfrac{5-\sqrt{5}}{2}\) ( loại )
\(x_2=\dfrac{5+\sqrt{5}}{2}\) ( nhận )
KL: \(x=\dfrac{5+\sqrt{5}}{2}\)
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