Question

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\frac{x+3}{2}\)

Answer 1

ĐK: \(x\ge1\)

\(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\frac{x+3}{2}\)

\(\Leftrightarrow\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}=\frac{x+3}{2}\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x-1}+1\right)^2}+\sqrt{\left(\sqrt{x-1}-1\right)^2}=\frac{x+3}{2}\)

\(\Leftrightarrow\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|=\frac{x+3}{2}\)

+) Xét \(x\ge2\):

\(pt\Leftrightarrow\sqrt{x-1}+1+\sqrt{x-1}-1=\frac{x+3}{2}\)

\(\Leftrightarrow4\sqrt{x-1}=x+3\)

\(\Leftrightarrow x=5\)(thỏa)

+) Xét \(1\le x< 2\):

\(pt\Leftrightarrow\sqrt{x-1}+1+1-\sqrt{x-1}=\frac{x+3}{2}\)

\(\Leftrightarrow x+3=4\)

\(\Leftrightarrow x=1\)(thỏa)

Vậy pt có 2 nghiệm \(x\in\left\{5;1\right\}\)