\(\Leftrightarrow4x^4+6x^2-2+x^2=0\)
\(\Leftrightarrow4x^4+5x^2-2=0\)(1)
Đặt \(x^2=a\left(a>=0\right)\)
(1) trở thành \(4a^2+5a-2=0\)(2)
\(\text{Δ}=5^2-4\cdot4\cdot\left(-2\right)=25+32=57>0\)
Do đó: Phương trình (2) có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}a_1=\dfrac{-5-\sqrt{57}}{8}\left(loại\right)\\a_2=\dfrac{-5+\sqrt{57}}{8}\left(nhận\right)\end{matrix}\right.\)
\(\Leftrightarrow x=\pm\sqrt{\dfrac{\sqrt{57}-5}{8}}\)
`2x^2(2x^2+3)=2-x^2`
`<=>4x^4+6x^2+x^2-2=0`
`<=>4x^4+7x^2-2=0`
Đặt `x^2=t (t >= 0)` khi đó ta có ptr:
`4t^2+7t-2=0`
`<=>4t^2-t+8t-2=0`
`<=>t(4t-1)+2(4t-1)=0`
`<=>(4t-1)(t+2)=0`
`<=>` $\left[\begin{matrix} t=\dfrac{1}{4} (t/m)\\ t=-2 (ko t/m)\end{matrix}\right.$
`@t=1/4=>x^2=1/4`
`<=>x=[+-1]/2`
Vậy `S={[+-1]/2}`
\(2x^2\left(2x^2+3\right)=2-x^2\)
Đặt \(x^2=a;a\ge0\)
`->` pt trở thành:
`<=>2a(2a+3)=2-a`
`<=>4a^2+7a-2=0`
\(\Leftrightarrow\left[{}\begin{matrix}a=\dfrac{1}{4}\left(tm\right)\\a=-2\left(ktm\right)\end{matrix}\right.\)
`=>`\(x=\pm\sqrt{\dfrac{1}{4}}=\pm\dfrac{1}{2}\)
Vậy \(S=\left\{\pm\dfrac{1}{2}\right\}\)
