a) Ta có: y = f(x) = 5 - 2x
+) f(0) => y = 5 - 2 . 0 = 5
+) f(-3) => y = 5 - 2 . (-3) = 5 - (-6) = 11
+) f(\(\frac{1}{4}\)) => y = 5 - 2 . \(\frac{1}{4}\) = \(5-\frac{1}{2}\) = 4,5
b)
+) Khi y = 5 => 5 = 5 - 2x
=> 2x = 5 - 5 = 0
=> x = 0
+) Khi y = 3 => 3 = 5 - 2x
=> 2x = 2
=> x = 1
+) Khi y = -1 => -1 = 5 - 2x
=> 2x = 6
=> x = 3
a) f(0)=5-2.0=5
f(-3)=5-2.(-3)=5-(-6)=5+6=11
f(1/4)=5-2.1/4=5-1/2=4/1/2