Question

(a+1)(a+2)(a+3)(a+6)>96 a²

Answer 1

Ответ:

(а+1)×(а+2)×(а+3)×(а+6)=35а >96а 2

Answer 2

To solve the inequality $(a+1)(a+2)(a+3)(a+6) > 96a^2$, you can follow these steps:

1. Expand both sides of the inequality:

$(a+1)(a+2)(a+3)(a+6) > 96a^2$

Expand the left side:

$a^4 + 12a^3 + 47a^2 + 72a + 36 > 96a^2$

2. Subtract $96a^2$ from both sides of the inequality to move all terms to one side:

$a^4 + 12a^3 + 47a^2 + 72a + 36 - 96a^2 > 0$

3. Simplify the equation:

$a^4 + 12a^3 - 49a^2 + 72a + 36 > 0$

4. Now, you need to find the roots of the equation $a^4 + 12a^3 - 49a^2 + 72a + 36 = 0$. These are the points where the inequality may change, so you'll want to determine the sign of the expression in each interval between these roots.

This equation may not have simple algebraic solutions for its roots, so you might need to use numerical methods or a graphing calculator to find approximate values.

Once you have the roots, you can test intervals between them to determine when the expression $a^4 + 12a^3 - 49a^2 + 72a + 36$ is positive. The solution to the inequality $(a+1)(a+2)(a+3)(a+6) > 96a^2$ will be the combination of these intervals where the expression is greater than zero.