Question

\(\sqrt{2x^4-8x^2+16}=\sqrt{8}.\)

Answer 1

\(ĐK:x\in R\)

\(PT\Leftrightarrow2x^4-8x^2+16=8\\ \Leftrightarrow x^4-4x^2+8=4\\ \Leftrightarrow x^4-4x^2+4=0\\ \Leftrightarrow\left(x^2-2\right)^2=0\\ \Leftrightarrow x^2-2=0\\ \Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)