Đặt \(\sqrt{x+\frac{1}{2}}=a\ge0\Rightarrow x=a^2-\frac{1}{2}\)
\(\sqrt{a^2-\frac{1}{2}+\frac{3}{4}+a}=a^2-\frac{1}{2}+\frac{5}{4}\)
\(\Leftrightarrow\sqrt{a^2+a+\frac{1}{4}}=a^2+\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}\sqrt{\left(2a+1\right)^2}=a^2+\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{2}\left(2a+1\right)=a^2+\frac{3}{4}\)
\(\Leftrightarrow4a^2-4a+1=0\)
\(\Rightarrow a=\frac{1}{2}\Rightarrow\sqrt{x+\frac{1}{2}}=\frac{1}{2}\Rightarrow x=-\frac{1}{4}\)