Question

Допоможіть, будь ласка! Даю 20балів 3 Доведіть, що: 1) ( x + + + ) ( x + + + ) 2 4. x 2) (x+1)(y+2)(2 + 8) ≥ 32 ху, якщо х 20, у 20, z 20. 24, якщо x > 0, y> 0; HE​

Answer 1

1.   (x + 2)(x + 4) = x^2 + 6x + 8 = x^2 + 6x + 8 = x^2 + 6x + 8, якщо x > 0.

2.   (x + 1)(y + 2)(z + 8) = xyz + 10xz + 16xy + 20x + 16y + 32z + 16. Якщо x > 0, y > 0, z > 0, то xyz, 10xz, 16xy, 20x, 16y, 32z всі будуть більші за нуль, тому (x + 1)(y + 2)(z + 8) також буде більше за 32.

Answer 2

1. (x^3 + 3x^2 + 3x + 1) (x^3 + 3x^2 + 3x + 1)

To expand this expression, we can use the binomial expansion formula for (a + b)^2: (a + b)^2 = a^2 + 2ab + b^2

Let's consider (x + 1) as 'a' and (x^2 + 2x + 1) as 'b':

(x + 1)^2 = x^2 + 2(x)(1) + 1^2 = x^2 + 2x + 1

Now, we can rewrite the original expression as follows:

(x^3 + 3x^2 + 3x + 1) (x^3 + 3x^2 + 3x + 1) = [(x^3 + 3x^2) + (3x + 1)] [(x^3 + 3x^2) + (3x + 1)] = (x^3 + 3x^2)^2 + 2(x^3 + 3x^2)(3x + 1) + (3x + 1)^2

Expanding further: = x^6 + 6x^5 + 9x^4 + 2(3x^4 + 9x^3 + 3x^3 + 9x^2) + 9x^2 + 6x + 1 = x^6 + 6x^5 + 9x^4 + 6(3x^4 + 6x^3 + 3x^2) + 9x^2 + 6x + 1 = x^6 + 6x^5 + 9x^4 + 18x^4 + 36x^3 + 18x^2 + 9x^2 + 6x + 1 = x^6 + 6x^5 + 27x^4 + 36x^3 + 27x^2 + 6x + 1

Therefore, (x^3 + 3x^2 + 3x + 1) (x^3 + 3x^2 + 3x + 1) simplifies to x^6 + 6x^5 + 27x^4 + 36x^3 + 27x^2 + 6x + 1.

2. (x + 1)(y + 2)(2x + 8) ≥ 32xy, if x > 0, y > 0

Let's substitute the given values x = 20, y = 20, z = 20 into the inequality:

(20 + 1)(20 + 2)(2(20) + 8) ≥ 32(20)(20) (21)(22)(48) ≥ 32(20)(20) 22896 ≥ 12800

Since 22896 is greater than or equal to 12800, the inequality holds true.

Therefore, (x + 1)(y + 2)(2x + 8) ≥ 32xy when x = 20, y = 20, z = 20, given that x > 0 and y > 0.