Ta có: \(\sqrt{\frac{1}{4}x^2+x+1}-\sqrt{6-2\sqrt{5}}=0\)
\(\Leftrightarrow\sqrt{\left(\frac{1}{2}x\right)^2+2\cdot\frac{1}{2}x\cdot1+1^2}-\sqrt{5-2\cdot\sqrt{5}\cdot1+1}=0\)
\(\Leftrightarrow\sqrt{\left(\frac{1}{2}x+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}=0\)
\(\Leftrightarrow\left|\frac{1}{2}x+1\right|-\left|\sqrt{5}-1\right|=0\)
\(\Leftrightarrow\left|\frac{1}{2}x+1\right|-\left(\sqrt{5}-1\right)=0\)
\(\Leftrightarrow\left|\frac{1}{2}x+1\right|=\sqrt{5}-1\)
\(\Leftrightarrow\left[{}\begin{matrix}\frac{1}{2}x+1=\sqrt{5}-1\\\frac{1}{2}x+1=1-\sqrt{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\frac{1}{2}x=\sqrt{5}-2\\\frac{1}{2}x=-\sqrt{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{5}-2}{2}\\x=-\frac{\sqrt{5}}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{\sqrt{5}-2}{2};-\frac{\sqrt{5}}{2}\right\}\)
