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Solve 3 > sqrt(3x)
3x >=o
x>=o/3
x>=0

square both sides and you have 3x < 3^2
3x,9
x<9/3
x<3

answer is 0<=x<3 correct or not?

3(cube) sqrt(2-5) = 3
cube both sides = 2-5x = 27
-5x = 25
x = -5, correct?

Thank you

1. Correct.

2. 3^3*sqrt(2 - 5x) = 3,
27*sqrt(2 - 5x) = 3,
Divide both sides by 27:
sqrt(2 - 5x) = 3/27 = 1/9,
Square both sides and get:
2 - 5x = 1/81,
-5x = 1/81 - 2,
-5x = 1/81 - 162/81,
-5x = -161/81,
Multiply both sides by -1/5:

x = (-161 / 81) * (-1/5) = 161 / 405.

CHECK:

3^3*sqrt(2 - (5*161/405)) =
27*sqrt(2 - 1.9877) =
27*sqrt(0.01235) =

27 * 0.11111 = 3.

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