ĐKXĐ:...
a/ \(\Leftrightarrow\sqrt{x^2+4\sqrt{x^2-4}}=16-2x^2\)
Đặt \(\sqrt{x^2-4}=a\ge0\Rightarrow x^2=a^2+4\)
\(\Leftrightarrow\sqrt{a^2+4+4a}=16-2\left(a^2+4\right)\)
\(\Leftrightarrow2a^2+a+2-8=0\)
\(\Leftrightarrow2a^2+a-6=0\) \(\Rightarrow\left[{}\begin{matrix}a=\frac{3}{2}\\a=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\sqrt{x^2-4}=\frac{3}{2}\Rightarrow x^2-4=\frac{9}{4}\)
b/
\(\Leftrightarrow\left(2x^2+1\right)\sqrt{2x^2+1}=2\left(2x^2+1\right)+2+3\sqrt{2x^2+1}\)
Đặt \(\sqrt{2x^2+1}=a>0\)
\(\Leftrightarrow a^3=2a^2+3a+2\)
\(\Leftrightarrow a^3-2x^2-3x-2=0\)
Nghiệm xấu, có lẽ bạn chép nhầm chỗ nào đó