\(\sqrt{2}\left(x^2+8\right)=5\sqrt{x^2+8}\)
\(\Leftrightarrow\sqrt{2\left(x^2+8\right)}=5\)
\(\Leftrightarrow2\left(x^2+8\right)=25\)
\(\Leftrightarrow2x^2=9\Leftrightarrow x^2=\frac{9}{2}\Leftrightarrow x=\pm\frac{3}{\sqrt{2}}\)
Hok tốt
\(\sqrt{2}\left(x^2+8\right)=5\sqrt{x^2+8}\)
\(\Rightarrow\left[\sqrt{2}\left(x^2+8\right)\right]^2=\left(5\sqrt{x^2+8}\right)^2\)
\(\Leftrightarrow2\left(x^4+16x^2+64\right)=25\left(x^2+8\right)\)
\(\Leftrightarrow2x^4+32x^2+128=25x^2+200\)
\(\Leftrightarrow2x^4+7x^2-72=0\)
\(\Leftrightarrow x^4+\frac{7}{2}x^2-36=0\)
\(\Leftrightarrow x^4+2.x^2.\frac{7}{4}+\frac{49}{16}-\frac{49}{16}-36=0\)
\(\Leftrightarrow\left(x^2+\frac{7}{4}\right)^2-\frac{625}{16}=0\)
\(\Leftrightarrow\left(x^2+\frac{7}{4}+\frac{25}{4}\right)\left(x^2+\frac{7}{4}-\frac{25}{4}\right)=0\)
\(\Leftrightarrow\left(x^2+8\right)\left(x^2-\frac{9}{2}\right)=0\left(1\right)\)
Ta thấy \(x^2\ge0;\forall x\)
\(\Rightarrow x^2+8\ge8>0;\forall x\left(2\right)\)
Từ (1) và (2) \(\Rightarrow x^2-\frac{9}{2}=0\)
\(\Leftrightarrow x^2=\frac{9}{2}\)
\(\Leftrightarrow x=\pm\frac{3}{\sqrt{2}}\)
Vậy tập hợp nghiệm của pt \(S=\left\{\frac{3}{\sqrt{2}};\frac{-3}{\sqrt{2}}\right\}\)
