Đặt \(x^2=t\left(t\ge0\right)\)
\(t^2-8t+16=0\)
=> \(=>t=4\)
\(x^2=t\\ =>\left[{}\begin{matrix}x=\sqrt{4}\\x=-\sqrt{4}\end{matrix}\right.\)
`<=> (x^2-4)^2 = 0`
`-> ((x-2)(x+2))^2 = 0`
`-> x = +-2`
`x^4 -8x^2 +16 =0`
`<=> (x^2)^2 - 2*x^2*4 + 4^2 =0`
`<=> (x^2 -4)^2 =0`
`=> x^2 -4 =0`
`=> x^2 =4`
`=> [(x=sqrt{4}=2),(x=-sqrt{4}=-2):}`
Vậy `S={2;-2}`
\(x^4-8x^2+16=0\)
\(\Leftrightarrow\left(x+2\right)^2\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)^2=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
Vậy \(S=\left\{2;-2\right\}.\)
