Bài 1:
\(f(2) = 0 \Rightarrow a.2 - 1 = 0 \Rightarrow 2a = 1\)\(\; \Rightarrow a = {1 \over 2}.\)
Bài 2: Ta có:
\(\;\;f\left( { - {1 \over 2}} \right) = \left( { - 2} \right).\left( { - {1 \over 2}} \right) + 3 = 4;\)
\(\eqalign{ & f\left( { - 5} \right) = \left( { - 2} \right).\left( { - 5} \right) + 3 = 13; \cr & f\left( 3 \right) = \left( { - 2} \right).3 + 3 = - 3. \cr} \)
Bài 3:
a) Ta có:
\(f\left( { - \sqrt 2 } \right) = - {\left( { - \sqrt 2 } \right)^2} + 2 \)\(\;= - 2 + 2 = 0;\)
\(f\left( {\sqrt 2 } \right) = - {\left( {\sqrt 2 } \right)^2} + 2 = - 2 + 2 = 0;\)
\(f\left( 3 \right) = - {\left( 3 \right)^2} + 2 = - 9 + 2 = - 7;\)
b) Ta có: \( - {x^2} + 2 = - 2 \Rightarrow - {x^2} = - 2 - 2 \)
\(\Rightarrow - {x^2} = - 4 \Rightarrow {x^2} = 4\)
\( \Rightarrow x = 2\) hoặc \(x = - 2\).
