\(\sqrt{\frac{x^2}{4}+\sqrt{x^2-4}}=8-x^2\)
\(\sqrt{\left(\frac{x^2-4+4\sqrt{x^2-4}+4}{4}\right)}=\sqrt{\left(\sqrt{x^2-4}+2\right)^2}=2\left(8-x^2\right)\)
Điều kiện: \(\left\{\begin{matrix}\left|x\right|\ge2\\\left|x\right|\le2\sqrt{2}\end{matrix}\right.\Rightarrow\left[\begin{matrix}-2\sqrt{2}\le x\le-2\\2\le x\le2\sqrt{2}\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{x^2-4}+2=2\left(8-x^2\right)\) đặt \(\sqrt{x^2-4}=t\)
\(\Leftrightarrow2\left(t^2-4\right)+t+2=0\Leftrightarrow2t^2+t-6=0\){delta =1+48=7^2}
\(\Rightarrow\left[\begin{matrix}t=\frac{-1-7}{4}\left(loiaj\right)\\t=\frac{-1+7}{4}=\frac{3}{2}\end{matrix}\right.\)
\(\Leftrightarrow x^2-4=\frac{9}{4}\Rightarrow x^2=\frac{25}{4}\Rightarrow\left[\begin{matrix}x=-\frac{5}{2}\\x=\frac{5}{2}\end{matrix}\right.\) nhận hết
