Question

3^(2х+1)+8•3^х -3≥0 Помогите, очень надо

Answer 1

3*3^(2x)+8*3^x-3≥0. Положим 3^x=t, t>0, тогда
3t²+8t-3≥0,     3(t+3)(t-1/3)≥0.
Поскольку t>0, то t+3>3, отсюда t-1/3≥0, т.е.3^x≥1/3, 3^x≥3^(-1),x≥-1, х∈[-1;+∞)

Answer 2

Solving the Inequality

To solve the inequality 3^(2x+1) + 8 * 3^x - 3 ≥ 0, we can follow these steps:

1. Rewrite the equation using exponent rules: 3^(2x+1) + 8 * 3^x - 3 ≥ 0. 2. Simplify the equation: 3^(2x+1) + 8 * 3^x ≥ 3. 3. Factor out a common term: 3^x(3^x + 8) ≥ 3.

Now, we have a quadratic inequality in terms of 3^x. To solve this inequality, we can consider two cases:

Case 1: 3^x > 0

If 3^x > 0, then we can divide both sides of the inequality by 3^x without changing the direction of the inequality:

3^x(3^x + 8) ≥ 3 3^x + 8 ≥ 3 / 3^x 3^x + 8 ≥ 1 / 3^(−x)

Case 2: 3^x < 0

If 3^x < 0, then we need to flip the direction of the inequality when dividing by 3^x:

3^x(3^x + 8) ≤ 3 3^x + 8 ≤ 1 / 3^(−x)

Now, let's solve each case separately.

Case 1: 3^x > 0

If 3^x > 0, we can divide both sides of the inequality by 3^x:

3^x + 8 ≥ 1 / 3^(−x) 3^x + 8 ≥ 3^(x)

Now, we have a quadratic inequality in terms of 3^x. To solve this, we can rewrite the inequality as:

3^(x) - 3^x + 8 ≥ 0 2 * 3^x + 8 ≥ 0 2 * 3^x ≥ -8 3^x ≥ -4

Since 3^x is always positive, the inequality 3^x ≥ -4 is always true. Therefore, there are no restrictions on the values of x in this case.

Case 2: 3^x < 0

If 3^x < 0, we need to flip the direction of the inequality when dividing by 3^x:

3^x + 8 ≤ 1 / 3^(−x) 3^x + 8 ≤ 3^(x)

Now, we have a quadratic inequality in terms of 3^x. To solve this, we can rewrite the inequality as:

3^(x) - 3^x + 8 ≤ 0 2 * 3^x + 8 ≤ 0 2 * 3^x ≤ -8 3^x ≤ -4

Since 3^x is always positive, the inequality 3^x ≤ -4 is never true. Therefore, there are no solutions in this case.

Conclusion

In conclusion, the inequality 3^(2x+1) + 8 * 3^x - 3 ≥ 0 has no restrictions on the values of x. Therefore, the solution set is (-∞, +∞).

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