Đặt \(t=x^2\)
\(\Leftrightarrow5t^2+4t-1=0\)
\(\Leftrightarrow\left(5t-1\right)\left(t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5t-1=0\\t+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=\dfrac{1}{5}\\t=-1\end{matrix}\right.\)
Với \(t=\dfrac{1}{5}\) :
\(\Leftrightarrow x^2=\dfrac{1}{5}\Leftrightarrow x=\pm\dfrac{1}{\sqrt{5}}\)
Với \(t=-1\)
\(\Leftrightarrow x^2=-1\Leftrightarrow x\in\varnothing\)
Vậy \(S=\left\{\pm\dfrac{1}{\sqrt{5}}\right\}\)
Đúng 0
Bình luận (0)
5x4 + 4x2 − 1 = 0
thế x = t2
=> 5t2 + 4t − 1 = 0
=> t =
=> t = −1
thay t = x2,ta có:
=> x2 =
=> x2 = −1
=> x =
=> x =
x ∉ R
x1 = ; x2 =
Đúng 0
Bình luận (0)
