ĐK: \(\left|x\right|\ge2\)
Đặt \(\frac{x^2}{4}=a\)
=> \(\left\{{}\begin{matrix}x^2=4a\\x^2-4=4\left(a-1\right)\end{matrix}\right.\)
Có : \(\sqrt{a+\sqrt{4\left(a-1\right)}}=8-4a\)
<=> \(\sqrt{a+2\sqrt{a-1}}=8-4a\)
<=> \(\sqrt{\left(\sqrt{a-1}+1\right)^2}=8-4a\)
<=> \(\sqrt{a-1}+1=8-4a\) <=> \(\sqrt{a-1}=7-4a\)<=> \(a-1=49-56a+16a^2\)
<=> \(0=16a^2-57a+50\) <=> \(\left(a-2\right)\left(a-\frac{25}{16}\right)=0\)
=> \(\left[{}\begin{matrix}a=2\\a=\frac{25}{16}\end{matrix}\right.\)<=> \(\left[{}\begin{matrix}\frac{x^2}{4}=2\\\frac{x^2}{4}=\frac{25}{16}\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x^2=8\\x^2=\frac{25}{4}\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=\pm2\sqrt{2}\left(ktm\right)\\x=\pm\frac{5}{2}\left(tm\right)\end{matrix}\right.\)
Vậy pt có tập nghiệm \(S=\left\{\pm\frac{5}{2}\right\}\)
