求这两道题的答案20

1、f(-1) = 2^1 = 2；f(0) = 2^0 = 1；f(1) = f(1-1) - f(1-2) = f(0) - f(-1) = 1 - 2 = -1 = -f(0)；f(2) = f(1) - f(0) = -f(0) - f(0) = -2f(0)；f(3) = f(2) - f(1) = -2f(0) + f(0) = -f(0)；f(4) = f(3) - f(2) = -f(0) + 2f(0) = f(0)；f(5) = f(4) - f(3) = f(0) + f(0) = 2f(0)；f(6) = f(5) - f(4) = 2f(0) - f(0) = f(0)；f(7) = f(6) - f(5) = f(0) - 2f(0) = -f(0)；f(8) = f(7) - f(6) = -f(0) - f(0) = -2f(0)；可见，f(6) = f(0)，f(7) = f(1)，……，f(x)是周期为6的周期函数。2020/6 = 336……4，所以f(2020) = f(3) = -f(0) = -1 。2、f(1) = f(-2)；f(2) = f(-1)，……，f(x)是周期为3的周期函数。f(5) = f(2) = f(-1) = (-1)^2 - 2^(-1) = 1 - 1/2 = 1/2 。