x=(x^2+3x-2)^2+3(x^2+3x-2)-2

x=(x^2+3x-2)^2+3(x^2+3x-2)-2

Let f (x) = x ^ 2 + 3x-2, then f (x ^ 2 + 3x-2) = X
Then f (f (x)) = X
x^2+3x-2=x
x^2+2x-2=0
x1=-1+√3
x2=-1-√3
You check it out, right,
I wish you study every day, come on!

(3x^2-x-4)/(2+x-x^2)≥x-2

It's more complicated. Wait a minute, I'll give you a slow answer. The original inequality is (3x ^ 2-x-4) / (2 + x-x ^ 2) - (X-2) ≥ 0, the general score is (3x ^ 2-x-4) / (2 + x-x ^ 2) - (2x + x ^ 2-x ^ 3-4-2x + 2x ^ 2) (2 + x-x ^ 2) ≥ 0, the general score is (3x ^ 2-x-4 + x ^ 3-3x ^ 2 + 4) / (2 + x-x ^ 2) ≥ 0, and then (x ^ 3-x) / (2 + x-x ^ 2) ≥ 0

How to solve the equation (x + 39) / (3x + 39 + 6) = 60%?

5(x+39)=3(3x+39+6)
5x+78=18+9x
4x=60
x=15
When x = 15, 3x + 39 + 6 is not equal to 0
x=15