Question

X(x + 6) + 8 <= 4(x + 2) x(12 - x) >= 12(x - 3)

Answer 1

Ответ:

1)x є [-2; 0]

2)x є [-6 ; 6]

Объяснение:

x(x + 6) + 8 <= 4(x + 2)

x^2 + 6x + 8 -4x - 8 <=0

x^2 + 2x <=0

x(x+2) <= 0

x є [-2; 0]

x(12 - x) >= 12(x - 3)

12x - x^2 -12x + 36>=0

x^2 - 36 <= 0

(x+6)(x-6) <= 0

x є [-6 ; 6]

Answer 2

Let's solve the given inequalities step by step:

1. X(x + 6) + 8 <= 4(x + 2):

Expanding both sides: X^2 + 6X + 8 <= 4X + 8

Rearranging the terms: X^2 + 6X - 4X <= 8 - 8

Simplifying: X^2 + 2X <= 0

Factoring: X(X + 2) <= 0

To solve this inequality, we need to consider the signs of the factors X and (X + 2). We have three cases:

Case 1: X < 0 If X is negative, both X and (X + 2) are negative or zero. In this case, the inequality is true.

Case 2: X = 0 If X is zero, both X and (X + 2) are zero. In this case, the inequality is true.

Case 3: X > 0 If X is positive, both X and (X + 2) are positive. In this case, the inequality is also true.

Therefore, the solution to the inequality X(x + 6) + 8 <= 4(x + 2) is X <= 0.

2. x(12 - x) >= 12(x - 3):

Expanding both sides: 12x - x^2 >= 12x - 36

Rearranging the terms: 0 >= x^2 - 36

Adding x^2 to both sides: x^2 <= 36

Taking the square root of both sides: |x| <= 6

To solve this inequality, we need to consider the signs of x. Since the expression is within absolute value, we have two cases:

Case 1: x >= 0 If x is non-negative, the absolute value is unnecessary, and the inequality becomes x <= 6. In this case, the inequality is true.

Case 2: x < 0 If x is negative, the absolute value changes the sign, and the inequality becomes -x <= 6. Multiplying both sides by -1, we get x >= -6. In this case, the inequality is also true.

Therefore, the solution to the inequality x(12 - x) >= 12(x - 3) is x <= 6 or x >= -6.

In summary:

1. X(x + 6) + 8 <= 4(x + 2) is true when X <= 0.
2. x(12 - x) >= 12(x - 3) is true when x <= 6 or x >= -6.