sif(x)=x^2-x,f(x+1)=f(-x)

解答

s i f (x) = x^{2} - x, f (x + 1) = f (- x)

s = 0, s \in R, s = 0, s \in R, x = 0, x = 0, x = 1, x = 1, f \in R f = 0 f \in R f = 0

求解步骤

s i f (x) = x^{2} - x

将

s i f (x)

改写成标准复数形式:

0 + s f (x) i

0 + s f (x) i = x^{2} - x

将

x^{2} - x

改写成标准复数形式:

(x^{2} - x) + 0 i

0 + s f (x) i = (x^{2} - x) + 0 i

复数仅在实部和虚部均相等时才相等改写为方程组:

[0 = x^{2} - x s f (x) = 0]

[0 = x^{2} - x s f (x) = 0] : s = 0, s \in R, s = 0, s \in R, x = 0, x = 0, x = 1, x = 1, f \in R f = 0 f \in R f = 0

s = 0, s \in R, s = 0, s \in R, x = 0, x = 0, x = 1, x = 1, f \in R f = 0 f \in R f = 0

(5 x + 4) + 7 i = (3 y + 1) i + 9

(x + i y)^{2} - x - i y + i (x - i y) = 0

(1 + 5 i) p - 2 q = 3 + 7 i

x + y i = 3 - 2 i

(x + i y)^{2} - 3 i (x + i y) - 2 = 0