Question

\(x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=2\)

Answer 1

ĐKXĐ: \(x\ge-\frac{1}{4}\)

\(\Leftrightarrow x+\sqrt{x+\frac{1}{4}+\sqrt{x+\frac{1}{4}}+\frac{1}{4}}=2\)

\(\Leftrightarrow x+\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=2\)

\(\Leftrightarrow x+\sqrt{x+\frac{1}{4}}+\frac{1}{2}=2\)

\(\Leftrightarrow\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2=2\)

\(\Rightarrow\sqrt{x+\frac{1}{4}}+\frac{1}{2}=\sqrt{2}\Rightarrow\sqrt{x+\frac{1}{4}}=\sqrt{2}-\frac{1}{2}\)

\(\Leftrightarrow x+\frac{1}{4}=\frac{9}{4}-\sqrt{2}\Rightarrow x=2-\sqrt{2}\)