Question

x4−2x2−3=0

Answer 1

\(x^4-2x^2-3=0\\ \Leftrightarrow x^4-3x^2+x^2-3=0\\ \Leftrightarrow x^2\left(x^2-3\right)+\left(x^2-3\right)=0\\ \Leftrightarrow\left(x^2+1\right)\left(x^2-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2+1=0\left(voli\right)\\x^2-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)

Answer 2

`x^4 -2x^2 -3=0`

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

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

\(< =>\left[{}\begin{matrix}x^2-1=2\\x^2-1=-2\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x^2=3\\x^2=-1\left(voli\right)\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)

Answer 3

Đặt t = x² (t ≥ 0)

Phương trình tương đương:

t² - 2t - 3 = 0

Ta có: a - b + c = 1 - (-2) - 3 = 0

Phương trình có hai nghiệm:

t₁ = -1 (loại)

t₂ = 3 (nhận)

Với t₂ = 3

⇔ x² = 3

⇔ x = √3; x = -√3

Vậy S = {-√3; √3}

Answer 4

    \(x^4-2x^2-3=0\)

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

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

\(\Leftrightarrow x^2\left(x^2-3\right)+\left(x^2-3\right)=0\)

\(\Leftrightarrow\left(x^2-3\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow x^2-3=0\)(do \(x^2+1>0\forall x\))

\(\Leftrightarrow x^2=3\)

\(\Leftrightarrow\)\(\left[\begin{array}{} x=\sqrt3\\ x=-\sqrt3 \end{array} \right.\)