Question

1)4x^2>=0 2)(x-1)(x-2)<0

Answer 1

1) 4x² ≥ 0

x² ≥ 0

Ответ : x ∈ (- ∞ ; + ∞)

2) (x - 1)(x - 2) < 0

     +           -              +

______₀______₀_____

            1             2

            //////////////

Ответ : x ∈ (1 ; 2)

Answer 2

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Готово!!Удачи))

Answer 3

1. The inequality 4x^2 >= 0 is true for all real values of x. In other words, any real number you substitute for x will satisfy this inequality. This is because when you square any real number, the result is always non-negative (greater than or equal to zero). Therefore, the solution set for this inequality is the set of all real numbers.

2. To solve the inequality (x - 1)(x - 2) < 0, we can use the method of sign chart or interval notation. Let's break it down into intervals based on the critical points, which are 1 and 2.

For x < 1: In this interval, both factors (x - 1) and (x - 2) are negative. A negative multiplied by a negative gives a positive, so the inequality is satisfied.

For 1 < x < 2: In this interval, (x - 1) is positive while (x - 2) is negative. A positive multiplied by a negative gives a negative, so the inequality is not satisfied.

For x > 2: In this interval, both factors (x - 1) and (x - 2) are positive. A positive multiplied by a positive gives a positive, so the inequality is satisfied.

Therefore, the solution to the inequality (x - 1)(x - 2) < 0 is x < 1 or 2 < x. In interval notation, the solution can be written as (-∞, 1) U (2, +∞).