Question

9^x - 1/9^x-1 - 1≥9 + 18/9^x - 3

Answer 1

Ответ:

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Answer 2

Let's simplify the given inequality step by step:

$9^x - \frac{1}{9^{x-1}} - 1 \geq 9 + \frac{18}{9^x} - 3$

First, combine the terms on the right side:

$9^x - \frac{1}{9^{x-1}} - 1 \geq \frac{27}{9^x} - 3$

Now, to simplify further, let's clear the fractions. Multiply both sides of the inequality by $9^{x+1}$ (the least common denominator):

$9^{x+1}(9^x - \frac{1}{9^{x-1}} - 1) \geq 9^{x+1}(\frac{27}{9^x} - 3)$

Simplify both sides:

$9^{2x+1} - 9^2 + 9^{x+1} \geq 27 - 3 \cdot 9^{x+1}$

Combine like terms:

$9^{2x+1} + 9^{x+1} - 81 \geq -18 \cdot 9^{x+1}$

Now, let's make a substitution to simplify further. Let $y = 9^x$, then $y^2 = (9^x)^2 = 9^{2x}$. Substituting $y$:

$y^2 + y - 81 \geq -18y$

Move all terms to one side of the inequality:

$y^2 + 19y - 81 \geq 0$

Now, let's factor the quadratic expression:

$(y - 3)(y + 27) \geq 0$

The solutions to this inequality are $y \leq -27$ or $y \geq 3$. Remember that $y = 9^x$, so we need to consider these solutions in the context of the original variable:

$9^x \leq -27 \quad \text{or} \quad 9^x \geq 3$

The first inequality ($9^x \leq -27$) has no real solutions because $9^x$ is always positive.

The second inequality ($9^x \geq 3$) is true for all $x$ because $9^x$ is always positive. Therefore, the solution to the original inequality is all real numbers $x$.