Đk \(2\le x\le6\)
Đặt \(\sqrt{2+x}+\sqrt{6-x}=t\left(t\ge0\right)\)
\(\Leftrightarrow\left(\sqrt{2+x}+\sqrt{6-x}\right)^2=t^2\)
\(\Leftrightarrow2+x+6-x+2\sqrt{\left(2+x\right)\left(6-x\right)}=t^2\)
\(\Leftrightarrow\sqrt{\left(2+x\right)\left(6-x\right)}=\dfrac{t^2-8}{2}\)
từ đó ta có pt theo biến t
\(t+\dfrac{t^2-8}{2}=8\)
\(\Leftrightarrow t^2+2t-24=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=4\left(nh\right)\\t=-6\left(l\right)\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{\left(2+x\right)\left(6-x\right)}=\dfrac{4^2-8}{2}=4\)
\(\Leftrightarrow\left(2+x\right)\left(6-x\right)=4^2\)
\(\Leftrightarrow12-2x+6x-x^2-16=0\)
\(\Leftrightarrow-4+4x-x^2=0\)
\(\Leftrightarrow x=2\)
