(ln^2 (x-1))*(4x^2-5x+1) >0

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вопрос

To solve this inequality, we need to find the critical points where the expression equals zero and where it changes sign.

First, set the expression equal to zero:
(ln^2(x-1))*(4x^2-5x+1) = 0

This can happen if either ln^2(x-1) = 0 or 4x^2-5x+1 = 0.

For ln^2(x-1) = 0, we have:
ln(x-1) = 0
x - 1 = e^0
x - 1 = 1
x = 2

For 4x^2-5x+1 = 0, we can factor it as:
(4x-1)(x-1) = 0

Setting each factor equal to zero gives:
4x-1 = 0
x = 1/4

x-1 = 0
x = 1

So the critical points are x = 1/4, 1, and 2.

Now, we plug in test points between the critical points into the inequality and check the sign:

For x < 1/4: choose x = 0:
(ln^2(0-1))(4(0)^2-5(0)+1) = (ln^2(-1))(1) = undefined, not applicable

For 1/4 < x < 1: choose x = 1/2:
(ln^2(1/2-1))(4(1/2)^2-5(1/2)+1) = (ln^2(-1/2))(1/2) = positive

For 1 < x < 2: choose x = 3/2:
(ln^2(3/2-1))(4(3/2)^2-5(3/2)+1) = (ln^2(1/2))(9-15/2+1) = negative

For x > 2: choose x = 3:
(ln^2(3-1))(4(3)^2-5(3)+1) = (ln^2(2))(36-15+1) = positive

Therefore, the inequality ln^2(x-1)*(4x^2-5x+1) > 0 holds for x ∈ (1/4,1) U (2,∞).