`3x^4+2x^2=16`
Đặt `x^2=t (t >=0)` có:
`3t^2+2t=16`
`<=>3t^2+2t-16=0`
`<=>` \(\left[ \begin{array}{l}t=2(TM)\\t=\dfrac{-8}{3}(L)\end{array} \right.\)
Với `t=2` : `x^2=2 <=> x= \pm \sqrt2`
Vậy `S={\pm \sqrt2}`.
Ta có: \(3x^4+2x^2=16\)
\(\Leftrightarrow3x^4+2x^2-16=0\)
\(\Leftrightarrow3x^4+6x^2-8x^2-16=0\)
\(\Leftrightarrow3x^2\left(x^2+2\right)-8\left(x^2+2\right)=0\)
\(\Leftrightarrow\left(x^2+2\right)\left(3x^2-8\right)=0\)
mà \(x^2+2>0\forall x\)
nên \(3x^2-8=0\)
\(\Leftrightarrow3x^2=8\)
\(\Leftrightarrow x^2=\dfrac{8}{3}\)
\(\Leftrightarrow x\in\left\{\dfrac{2\sqrt{6}}{3};-\dfrac{2\sqrt{6}}{3}\right\}\)
Vậy: \(S=\left\{\dfrac{2\sqrt{6}}{3};-\dfrac{2\sqrt{6}}{3}\right\}\)
