Question

ПОМОГИТЕ ПОМОГИТЕ!!!! НЕ могу решить!!!

Answer 1

1)
(5/6)^x * (4/5)^x = (8/15)*(5/6)
[(5/6)*(4/5)]^x = 4/9
(2/3)^x = (2/3)^2
x = 2

2)
3*(81)^(1/x) - 10*9^(1/x) + 3 = 0
3*(3^4)^(1/x) - 10*(3^2)^(1/x) + 3 = 0
3*3^(4/x) - 10*3^(2/x) + 3 = 0
Сделаем подстановку
y = 3^(2/x)
3*y^2 - 10*y +3 = 0
Получилось квадратное уравнение относительно y
a=3, b=-10, c=3
D=(-10)^2 - 4*3*3 = 100 - 36 = 64, D>0, два корня.
D^(1/2) = 64^(1/2) = 8
y1 = (10 - 8) / (2*3) = 2/6 = 1/3
y2 = (10+8) / (2*3) = 18/ 6 = 3

Возвращаемся к переменной x
1/3 =  3^(2/x)
3^(-1) = 3^(2/x)
-1 = 2/x
x1= - 2

3 =  3^(2/x)
3^1 = 3^(2/x)
1 = 2/x
x2 = 2

Проверка
3*(81)^(-1/2) - 10*(9)^(-1/2) +3 = 0
3/81^(1/2) - 10/9^(1/2) + 3 = 0
3/9 - 10/3 + 3 = 0
1/3 - 10/3 + 9/3 = 0
(1-10+9)/3 = 0
0/3 = 0
0 = 0
x1 подходит
3*(81)^(1/2) - 10*(9)^(1/2) +3 = 0
3*9 -10*3 + 3 = 0
27 - 30 +3 = 0
30-30 = 0
0 = 0
x2 подходит