sif(x)=-2x+5,E(s,x)=f(1)+f(-1)

解答

s i f (x) = - 2 x + 5, E (s, x) = f (1) + f (- 1)

(s = 0, s \in R, x = \frac{5}{2}, x = \frac{5}{2}, f \in R f = 0)

求解步骤

s i f (x) = - 2 x + 5

将

s i f (x)

改写成标准复数形式:

0 + s f (x) i

0 + s f (x) i = - 2 x + 5

将

- 2 x + 5

改写成标准复数形式:

(- 2 x + 5) + 0 i

0 + s f (x) i = (- 2 x + 5) + 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 = - 2 x + 5 s f (x) = 0]

[0 = - 2 x + 5 s f (x) = 0] : (s = 0, s \in R, x = \frac{5}{2}, x = \frac{5}{2}, f \in R f = 0)

(s = 0, s \in R, x = \frac{5}{2}, x = \frac{5}{2}, f \in R f = 0)

3 z^{2} + (2 - 3 i) z - 1 - 3 i = 0

r (t) = (2 t + 1) i + (t^{2} - 1) j

2 x + 51 = 4 (2 + y i) - 12

z^{2} - (1 + 9 i) z - 20 + 5 i = 0

z = z (a + i)