Đáp án:
$\begin{array}{l}
a)\\
+ f\left( 1 \right) = - {2.1^2} = - 2\\
+ f\left( { - 2} \right) = - 2.{\left( { - 2} \right)^2} = - 8\\
+ f\left( {\frac{{ - \sqrt 2 }}{2}} \right) = - 2.{\left( {\frac{{ - \sqrt 2 }}{2}} \right)^2} = - 2.\frac{1}{2} = - 1\\
b)y = 4.\left( {\sqrt 3 - 2} \right)\\
\Rightarrow - 2.{x^2} = 4.\left( {\sqrt 3 - 2} \right)\\
\Rightarrow {x^2} = - 2.\left( {\sqrt 3 - 2} \right) = 4 - 2\sqrt 3 \\
\Rightarrow {x^2} = {\left( {\sqrt 3 - 1} \right)^2}\\
\Rightarrow \left[ \begin{array}{l}
x = \sqrt 3 - 1\\
x = 1 - \sqrt 3
\end{array} \right.\\
Vậy\,khi:y = 4\left( {\sqrt 3 - 2} \right)\,thì:x = \sqrt 3 - 1\,hoặc\,x = 1 - \sqrt 3
\end{array}$
