a) \(f\left(0\right)\Rightarrow y=\left|0^2-1\right|=\left|-1\right|=1\)
\(f\left(-1\right)\Rightarrow y=\left|\left(-1\right)^2-1\right|=\left|1-1\right|=\left|0\right|=0\)
\(f\left(1\right)\Rightarrow y=\left|1^2-1\right|=\left|1-1\right|=\left|0\right|=0\)
\(f\left(\dfrac{1}{2}\right)\Rightarrow y=\left|\left(\dfrac{1}{2}\right)^2-1\right|=\left|\dfrac{1}{4}-1\right|=\left|\dfrac{-3}{4}\right|=\dfrac{3}{4}\)
b) ta có : \(y=7\Rightarrow\left|x^2-1\right|=7\Leftrightarrow\) \(\left\{{}\begin{matrix}x^2-1=7\\x^2-1=-7\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=8\\x^2=-6\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x=\sqrt{8}\\x=-\sqrt{8}\end{matrix}\right.\\x\in\varnothing\end{matrix}\right.\) vậy \(x=\sqrt{8};x=-\sqrt{8}\)
ta có : \(y=17\Leftrightarrow\left|x^2-1\right|=17\Leftrightarrow\) \(\left\{{}\begin{matrix}x^2-1=17\\x^2-1=-17\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2=18\\x^2=-16\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left\{{}\begin{matrix}x=\sqrt{18}\\x=-\sqrt{18}\end{matrix}\right.\\x\in\varnothing\end{matrix}\right.\) vậy \(x=\sqrt{18};x=-\sqrt{18}\)
