Bài 1:
a) Ta có: \(f\left( { - 2} \right) = 3 - \left( { - 2} \right) = 5;\)
\(f\left( { - 1} \right) = 3 - \left( { - 1} \right) = 4\);
\(f\left( 0 \right) = 3 - \left( 0 \right) = 3;\)
\(f\left( {{1 \over 2}} \right) = 3 - \left( {{1 \over 2}} \right) = {5 \over 2}.\)
b) Ta có:
\(\;\;3 - x = 5 \Rightarrow x = 3 - 5 \Rightarrow x = - 2;\)
\(\eqalign{ & 3 - x = 2 \Rightarrow x = 3 - 2 \Rightarrow x = 1; \cr & 3 - x = - 1 \Rightarrow x = 3 + 1 \Rightarrow x = 4. \cr} \)
Bài 2: Ta có: \(f\left( 1 \right) = - 1 \Rightarrow 3.1 + m = - 1 \)
\(\Rightarrow m = - 1 - 3 = - 4\)
Bài 3:
\(f\left( 0 \right) = - 3 \Rightarrow a.0 + b = - 3 \Rightarrow b = - 3\)
Vậy f(x) = ax – 3.
Lại có : \(f( - 1) = - 5 \Rightarrow a.( - 1) - 3 = - 5 \)
\(\Rightarrow - a = 3 - 5 \Rightarrow a = 2.\)
Vậy: \(f\left( x \right) = 2x - 3.\)
