C3 (2x^2-2x+1)/(2x-1)<=1 25x^2-3|3-5x|<30x-9

Предыдущий
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вопрос

To solve the inequality (2x^2 - 2x + 1)/(2x - 1) <= 1, we first need to find the critical points where the numerator and denominator become zero.

Setting the numerator equal to zero gives us:
2x^2 - 2x + 1 = 0

Using the quadratic formula, we find that the solutions are x = 1/2.

Setting the denominator equal to zero gives us:
2x - 1 = 0
x = 1/2

Now, we can determine the sign of the expression (2x^2 - 2x + 1)/(2x - 1) in each interval:
1) x < 1/2: both numerator and denominator are positive, so the expression is positive.
2) x > 1/2: both numerator and denominator are positive, so the expression is positive.

Therefore, the solution to the inequality is x belongs to (-∞, ∞).

For the inequality 25x^2 - 3|3 - 5x| < 30x - 9, we can start by breaking it down into separate cases based on the absolute value expression:

Case 1: 3 - 5x >= 0
25x^2 - 3(3 - 5x) < 30x - 9
25x^2 - 9 + 15x < 30x - 9
25x^2 - 15x - 9 < 30x - 9
25x^2 - 45x < 0
5x(5x - 9) < 0

This gives us two critical points: x = 0 and x = 9/5.

Case 2: 3 - 5x < 0
25x^2 - 3(5x - 3) < 30x - 9
25x^2 - 15x + 9 < 30x - 9
25x^2 - 45x + 18 < 30x - 9
25x^2 - 75x + 27 < 0

This inequality holds true for all x in the interval (-∞, ∞).

Therefore, the overall solution to the inequality 25x^2 - 3|3 - 5x| < 30x - 9 is x belongs to (-∞, 9/5).