Найдите наименьший положительный период функций: 1) 2sin0,25x+sinπ/2

Question

Алгебра: Найдите наименьший положительный период функций: 1) 2sin0,25x+sinπ/2

Anastasiya2108 · Accepted Answer

1) = -5(х +0,2)(х - 1)
D= 16 - 4(-5) = 36
х1 = (-4+6) / (-10) = -0,2
х2 = (-4 - 6) / (-10) = 1

2) = -21(х + 2/7)(х - 2/6) = - 3*7(х + 2/7)(х - 2/6) =
= - (7х + 2)(3х - 1)
D= 1 - 4(-21)*2 = 169
√D= √169 = 13
х1 = (-1+13 ) / (-42) = - 12/42 = - 2/7
х2 = (-1-13 ) / (-42) = 14/42 = 2/6

3) = -12(х - 1/3)(х - 1/4) = - (3х - 1)(4х - 1)
D= 49 - 4(-12)(-1) = 49 - 48 = 1
х1 = (-7+1) / (-24) = 6/24 = 1/4
х2 = (-7-1) / (-24) = 8/24 = 1/3

4) = 27(х - 1/3)(х + 1/9) = 3*9*(х - 1/3)(х + 1/9) =
= (3х - 1)(9х + 1)
D= 36 - 4*27*(-1) = 36 + 108 = 144
х1 = (6+12) / 54 = 18/54 = 1/3
х2 = (6-12) / 54 = -6/54 = - 1/9

MOGZ ответил