Question

2^х²+3х - 8 × 2^х> 0​

Answer 1

$> 0\\2^{x^{2} + 3x} - 2^{3} * 2^{x} > 0\\2^{x^{2} + 3x} - 2^{x + 3} > 0\\2^{x^{2} + 3x} > 2^{x + 3} => x^{2} + 3x > x + 3 => x^{2} + 2x - 3 > 0\\\\x^{2} + 2x - 3 = 0 = > x1 = 1; x2 = -3$

Ответ: x ∈ (-∞ , -3) ∪ (1 , ∞)

Answer 2

The inequality you provided is: 2^(x^2 + 3x) - 8 * 2^x > 0.

To solve this inequality, let's break it down into parts.

First, we can simplify the inequality by factoring out 2^x from both terms:

2^x * (2^(x^2 + 3x - 1) - 8) > 0.

Now we have two factors: 2^x and (2^(x^2 + 3x - 1) - 8). For the inequality to be true, either both factors must be positive or both must be negative.

Case 1: Both factors are positive. In this case, we have two possibilities:

1. 2^x > 0 and 2^(x^2 + 3x - 1) - 8 > 0. Since 2^x is always positive, we only need to consider the second inequality: 2^(x^2 + 3x - 1) - 8 > 0. Unfortunately, this inequality cannot be easily solved algebraically. We would need to use numerical methods or graphing to determine the specific values of x that satisfy it.

2. 2^x < 0 and 2^(x^2 + 3x - 1) - 8 < 0. Since 2^x is always positive, this case is not possible.

Case 2: Both factors are negative. In this case, we have two possibilities:

1. 2^x < 0 and 2^(x^2 + 3x - 1) - 8 < 0. Since 2^x is always positive, this case is not possible.

2. 2^x > 0 and 2^(x^2 + 3x - 1) - 8 > 0. In this case, we need to solve the inequality 2^(x^2 + 3x - 1) - 8 > 0. Similarly to Case 1, this inequality cannot be easily solved algebraically and requires numerical or graphical methods.

In conclusion, the inequality 2^(x^2 + 3x) - 8 * 2^x > 0 cannot be easily solved algebraically. Additional methods are needed to find the specific values of x that satisfy the inequality.