\(\sqrt{\left(\dfrac{1}{2}x+1\right)^2}\)-\(\sqrt[]{\left(\sqrt{5}-1\right)^2}\)=0
=>1/2x+1-\(\sqrt{5}\)+1=0(x>-2)
=>1/2x+2-\(\sqrt{5}\)=0
=>................
\(\sqrt{\dfrac{1}{4}x^2+x+1}-\sqrt{6-2\sqrt{5}}=0\Leftrightarrow\sqrt{\left(\dfrac{1}{2}x+1\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}=0\Leftrightarrow\left|\dfrac{1}{2}x+1\right|=\sqrt{5}-1\)\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+1=\sqrt{5}-1\\\dfrac{1}{2}x+1=1-\sqrt{5}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-4+2\sqrt{5}\\x=-2\sqrt{5}\end{matrix}\right.\)
Ta có: \(\sqrt{\dfrac{1}{4}x^2+x+1}-\sqrt{6-2\sqrt{5}}=0\)
\(\Leftrightarrow\left|\dfrac{1}{2}x+1\right|=\sqrt{5}-1\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x+1=\sqrt{5}-1\left(x\ge-2\right)\\\dfrac{1}{2}x+1=-\sqrt{5}+1\left(x< -2\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}x=\sqrt{5}-2\\\dfrac{1}{2}x=-\sqrt{5}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-4+2\sqrt{5}\left(nhận\right)\\x=-2\sqrt{5}\left(nhận\right)\end{matrix}\right.\)
