Question

Phương trình trùng phương :

`-x^4 - 2x^2 + 3 = 0`

Answer 1

`-x^4 - 2x^2 + 3 = 0`

`<=> -( x^4 +2.x^2 +1 -1) +3 = 0`

`<=> -(x^2+1)^2+4=0`

`<=>(2 -x^2 -1).(2 +x^2 +1) = 0`

`<=>(1-x).(1+x).(3+x^2) = 0 `

$\left[\begin{matrix} 1-x=0\\ 1+x=0\end{matrix}\right.$

$\left[\begin{matrix} x=1\\ x=-1\end{matrix}\right.$

Answer 2

\(-x^4-2x^2+3=0\)

`<=>x^4+2x^2-3=0`

\(\Delta=2^2-4.\left(-3\right)=4+12=16>0\)

\(\rightarrow\left\{{}\begin{matrix}x^2=\dfrac{-2+\sqrt{16}}{2}=1\left(tm\right)\\x^2=\dfrac{-2-\sqrt{16}}{2}=-3\left(ktm\right)\end{matrix}\right.\)

\(\Leftrightarrow x^2=1\)

\(\Rightarrow x=\pm1\)

Vậy \(S=\left\{\pm1\right\}\)