Question

Giải phương trình: \(x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{2}}}=3\)

Answer 1

đề bài là: \(x+\sqrt{x+\frac{1}{2}+\sqrt{x+\frac{1}{4}}}=3\)

Answer 2

ĐKXĐ: \(x\ge-\frac{1}{4}\)

\(x+\sqrt{x+\frac{1}{4}+2.\frac{1}{2}\sqrt{x+\frac{1}{4}}+\left(\frac{1}{2}\right)^2}=3\)

\(\Leftrightarrow x+\sqrt{\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2}=3\)

\(\Leftrightarrow x+\sqrt{x+\frac{1}{4}}+\frac{1}{2}=3\)

\(\Leftrightarrow x+\frac{1}{4}+\sqrt{x+\frac{1}{4}}+\frac{1}{4}=3\)

\(\Leftrightarrow\left(\sqrt{x+\frac{1}{4}}+\frac{1}{2}\right)^2=3\)

\(\Leftrightarrow\sqrt{x+\frac{1}{4}}+\frac{1}{2}=\sqrt{3}\) (do \(\sqrt{x+\frac{1}{4}}+\frac{1}{2}>0\) với mọi x)

\(\Leftrightarrow\sqrt{x+\frac{1}{4}}=\sqrt{3}-\frac{1}{2}\)

\(\Leftrightarrow x+\frac{1}{4}=\left(\sqrt{3}-\frac{1}{2}\right)^2\)

\(\Leftrightarrow...\)