Question

Доведіть що за будь-яких значень m і n нерівність є правильною: 1) (1+m)²>(m+6)(m-4);2) 30mn≤9m²+25n². помогите пожалуйста!! очень срочно!​

Answer 1

Ответ:

Пошаговое объяснение:

1) (1+m)²>(m+6)(m-4)

1+2m+m²>m²+6m-4m-24

1+2m+m²>m²+2m-24

1+2m+m²-(m²+2m-24)>0

1+2m+m²-m²-2m+24>0

25>0    это верно!

2) 30mn≤9m²+25n²

9m²+25n²-30mn≥0

(3m)²-2·(3m)·(5n)+(5n)²≥0

(3m-5n)²≥0    это верно!

квадрат любого числа ≥0

Answer 2

1) (1+m)² > (m+6)(m-4)

To prove that the inequality is true for any values of m and n, we need to show that it holds for all possible values. Let's start by expanding both sides of the inequality:

Left side: (1+m)² = 1 + 2m + m²

Right side: (m+6)(m-4) = m² + 2m - 24

Now, let's compare the expanded expressions:

1 + 2m + m² > m² + 2m - 24

Simplifying the inequality:

1 > -24

This inequality is true for any value of m. Therefore, the inequality (1+m)² > (m+6)(m-4) is true for all values of m.

2) 30mn ≤ 9m² + 25n²

To prove that this inequality is true for any values of m and n, we need to show that it holds for all possible values. Let's analyze the inequality:

30mn ≤ 9m² + 25n²

To simplify the inequality, let's subtract 30mn from both sides:

0 ≤ 9m² + 25n² - 30mn

Now, let's factor the right side of the inequality:

0 ≤ (3m - 5n)²

Since the square of any real number is always non-negative, the inequality 0 ≤ (3m - 5n)² is true for any values of m and n.

Therefore, the inequality 30mn ≤ 9m² + 25n² is true for all values of m and n.

Please let me know if you need any further clarification or assistance!