\(x^4-x^2-20=0\)
\(\Leftrightarrow x^4-4x^2+5x^2-20=0\)
\(\Leftrightarrow x^2\left(x^2-4\right)+5\left(x^2-4\right)=0\)
\(\Leftrightarrow\left(x^2-4\right)\cdot\left(x^2+5\right)=0\)
Nhận thấy; \(x^2+5>0\forall x\)
\(\Rightarrow x^2-4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=-2\end{matrix}\right.\)
Giải:
\(x^4-x^2-20=0\)
\(\Leftrightarrow-20-x^2+x^4=0\)
\(\Leftrightarrow\left(-4-x^2\right)\left(5-x^2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}-4-x^2=0\\5-x^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2=-4\\x^2=5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\pm\sqrt{5}\end{matrix}\right.\)
Vậy ...
Chúc bạn học tốt!
x4-x2-20=0
x4-4x2+5x2-20=0
x2(x2-4)+5(x2-4)=0
(x2-4)(x2+5)=0
Suy ra x2-4=0 hoặc x2+5=0
x2 =4 x2 =-5
x =-2 hoặc 2 x =+-\(\sqrt{5}\)
vậy x=+-2 hoặc x=+-\(\sqrt{5}\)
