Решить уравнения (x3+1)(x3++1)(x3-1)=44 (5x+1)(5x-+1)(5x+1)=48 - вопрос №3131314451515148414141509238066 от yakubovmalik 15.06.2021 18:26

Question

Алгебра: Решить уравнения (x3+1)(x3++1)(x3-1)=44 (5x+1)(5x-+1)(5x+1)=48 (4x+1)(4x++1)(4x-1)=50 заранее

yakubovmalik · Answer

1) (x³ + 1)(x³ + 1) - (x³ + 1)(x³ - 1) = 44
(x³ + 1)(x³ + 1 - x³ + 1) = 44
2(x³ + 1) = 44 |:2
x³ + 1 = 22
x³ = 21
x = $\sqrt[3]{21}$

Если тройки являются коэффициентом, а не степнем, то:
(3x + 1)(3x + 1) - (3x + 1)(3x - 1) = 44
(3x + 1)(3x + 1 - 3x + 1) = 44
2(3x + 1) = 44 |:2
3x + 1 = 22
3x = 21 |:3
x = 7

2) (5x + 1)(5x - 1) - (5x + 1)(5x + 1) = 48
(5x + 1)(5x - 1 - 5x - 1) = 48
-2(5x + 1) = 48 |: -2
5x + 1 = -24
5x = -25 |:5
x = -5

3) (4x + 1)(4x + 1) - (4x + 1)(4x - 1) = 50
(4x + 1)(4x + 1 - 4x + 1) = 50
2(4x + 1) = 50 |:2
4x + 1 = 25
4x = 24 |:4
x = 6