Question

Giải pt:

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

Answer 1

ĐKXĐ: \(x\ge\frac{1}{2}\)

Ta có: \(\sqrt{x+\sqrt{2x-1}}+\sqrt{x-\sqrt{2x-1}}=\sqrt{2}\)

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

\(\Leftrightarrow2x+\sqrt{x^2-\left(2x-1\right)}=2\)

\(\Leftrightarrow2x+\sqrt{x^2-2x+1}=2\)

\(\Leftrightarrow2x+\sqrt{\left(x-1\right)^2}=2\)

\(\Leftrightarrow2x+\left|x-1\right|=2\)(1)

Trường hợp 1: \(\frac{1}{2}\le x< 1\)

(1)\(\Leftrightarrow2x+1-x=2\)

\(\Leftrightarrow x+1=2\)

hay x=1(loại)

Trường hợp 2: \(x\ge1\)

(1)\(\Leftrightarrow2x+x-1=2\)

\(\Leftrightarrow3x-1=2\)

\(\Leftrightarrow3x=3\)

hay x=1(nhận)

Vậy: S={1}