Question

пожалуйста полностью с решением
даю 30 баллов

Answer 1

1)
root2 (x) + root4 (x) - 6 = 0
root4 (x) = t
t^2 + t - 6 = 0
По теореме Виета: t1 = -3, t2 = 2
root4 (x) = -3 => нет решений
root4 (x) = 2 => x = 2^4 = 16
x = 16

2)
root3 (x) + root6 (x) - 2 = 0
root6 (x) = t
t^2 + t - 2 = 0
По теореме Виета: t1 = -2, t2 = 1
root6 (x) = -2 => нет решений
root6 (x) = 1 => x = 1^6 = 1
x = 1

3)
root2 (x) - 3 root4 (x) - 10 = 0
root4 (x) = t
t^2 -3t - 10 = 0
По теореме Виета: t1 = -2, t2 = 5
root4 (x) = -2 => нет решений
root4 (x) = 5 => x = 5^4 = 625
x = 625

4)
root3 (x) - 3 root6 (x) - 18 = 0
root6 (x) = t
t^2 - 3t - 18 = 0
По теореме Виета: t1 = -3, t2 = 6
root6 (x) = -3 => нет решений
root6 (x) = 6 => x = 6^6
x = 6^6

Answer 2

Да