Question

(2х-3)²>=(3х-2)² решите​

Answer 1

ответ [-1;1] то есть от - 1 до 1

Answer 2

Ответ:

x ∈ [-1;1]

Пошаговое объяснение:

Раскроем скобки по формуле квадрат разности

4x^2-12x+9>=9x^2-12x+4

Приводим подобные слагаемые

5x^2 <= 5

x^2 <= 1

x^2-1<=0

По формуле разность квадратов

(x-1)*(x+1)<=0 => нули функции в точках -1 ; 1

Строим координатную прямую и берем промежуток ,где значения получаются отрицательные,т.к. нам надо меньше или равно нуля,получаем: x ∈ [-1;1]

Answer 3

To solve the inequality (2x - 3)² ≥ (3x - 2)², we need to consider two cases: when the expression inside the square (the base) is positive or zero, and when it is negative.

Case 1: Base is non-negative (when the base is zero or positive) (2x - 3)² ≥ 0, and (3x - 2)² ≥ 0

Since both squares are non-negative for all real values of x, the inequality holds true for this case.

Case 2: Base is negative (when the base is negative) For (2x - 3)² ≥ (3x - 2)² to hold true when the base is negative, we need to be cautious about the signs of the bases.

(2x - 3)² ≥ (3x - 2)²

Taking the square root of both sides (and considering the positive and negative roots):

|2x - 3| ≥ |3x - 2|

Now, we need to consider two sub-cases based on the signs of (2x - 3) and (3x - 2):

Sub-case 2.1: (2x - 3) is positive and (3x - 2) is positive: In this case, we don't need the absolute value signs:

(2x - 3) ≥ (3x - 2)

Now, solve for x:

2x - 3 ≥ 3x - 2

Subtract 2x from both sides:

-3 ≥ x - 2

Add 2 to both sides:

-1 ≥ x

Sub-case 2.2: (2x - 3) is negative and (3x - 2) is negative: In this case, we have to flip the inequality when removing the absolute value signs:

-(2x - 3) ≥ -(3x - 2)

Simplify:

-2x + 3 ≥ -3x + 2

Add 3x to both sides:

x + 3 ≥ 2

Subtract 3 from both sides:

x ≥ -1

So, for sub-case 2.2, the solution is x ≥ -1.

Combining the results from both sub-cases:

x ≤ -1 or x ≥ -1

Thus, the solution to the inequality (2x - 3)² ≥ (3x - 2)² is x ≤ -1 or x ≥ -1.