a)Ta có : \(y=f\left(x\right)=2x^2-\dfrac{1}{4}\)
\(\Rightarrow y=f\left(0\right)=2.0^2-\dfrac{1}{4}=\dfrac{-1}{4}\)
\(f\left(\dfrac{-1}{2}\right)=2.\left(\dfrac{-1}{2}\right)^2-\dfrac{1}{4}=2.\dfrac{1}{4}-\dfrac{1}{4}=\dfrac{1}{4}\)
b)Ta có : \(\dfrac{27}{5}=2x^2-\dfrac{1}{4}\)
\(\Rightarrow2x^2=\dfrac{27}{5}+\dfrac{1}{4}=\dfrac{113}{20}\)
\(\Rightarrow x^2=\dfrac{113}{40}\)
\(\Rightarrow x=\sqrt{\dfrac{113}{40}}\)
Sửa đề: \(y=f\left(x\right)=2x^2-\dfrac{1}{4}\)
a)\(\Rightarrow\left\{{}\begin{matrix}f\left(0\right)=2.0^2-\dfrac{1}{4}=0-\dfrac{1}{4}=-\dfrac{1}{4}\\f\left(-\dfrac{1}{2}\right)=\left(-\dfrac{1}{2}\right)^2-\dfrac{1}{4}=\dfrac{1}{4}-\dfrac{1}{4}=0\end{matrix}\right.\)
b) \(y=\dfrac{27}{5}\Leftrightarrow2x^2-\dfrac{1}{4}=\dfrac{27}{5}\)
\(\Leftrightarrow2x^2=\dfrac{27}{5}+\dfrac{1}{4}=\dfrac{113}{20}\Leftrightarrow x^2=\dfrac{113}{40}\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{113}{40}}\\x=-\sqrt{\dfrac{113}{40}}\end{matrix}\right.\)
