Question

3x^4+2x^2=16

Answer 1

`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}`.

Answer 2

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: \(S=\left\{\dfrac{2\sqrt{6}}{3};-\dfrac{2\sqrt{6}}{3}\right\}\)