[mathematics of senior one] if f (x) satisfies f (x) = - f (x + 3 / 2) and f (- 2) = f (- 1) = - 1, f (0) = 2, then f (1) + F (2) +. F (2009) =?

[mathematics of senior one] if f (x) satisfies f (x) = - f (x + 3 / 2) and f (- 2) = f (- 1) = - 1, f (0) = 2, then f (1) + F (2) +. F (2009) =?

Because f (x) = - f (x + 3 / 2), f (x) = f (x + 3), so f (x) is a periodic function with a period of 3, and f (- 2) = f (- 1) = - 1, that is, f (1) = f (- 2) = - 1F (2) = f (- 1) = - 1F (0) = 2, so f (0) + F (1) + F (2) = - 1 + (- 1) + 2 = 0 and 2009 / 3 = 669 2 so f (1) + F (2) +. F (2009) = f (1) + F (2) = -

Simplify the evaluation; X (X-2) - (x + 1) (x-3) where x = 5
(- 3A & # 179;) & # 178; × A & # 179; + (- 4A & # 178;) × a seventh power + (- 5A & # 179;) & # 179;

1.　　x(x--2)--(x+1)(x--3)
=(x^2--2x)--(x^2--2x--3)
=x^2--2x--x^2+2x+3
=3.
The value of this algebra is independent of X
　　2.(--3a^3)^2xa^3+(--4a^2)xa^7+(--5a^3)^3
=9a^6xa^3+(--4a^9)+(--125a^9)
=9a^9--4a^9--125a^9
=--120a^9.