Question

(5^x -5)*(13^x +2):(7^x -1)>=0

Answer 1

Поскольку - на - = +, то подходит всё кроме 0 (7 в нулевой степени 1)
Поэтому ответ (-∞;0)U[1;+∞)
$> </div> <div class=$ 0 0

Answer 2

To solve the inequality $(5^x - 5) \cdot (13^x + 2) : (7^x - 1) \geq 0$, we need to find the intervals for $x$ that satisfy this inequality.

Step 1: Identify the critical points The inequality is only undefined when the denominator $(7^x - 1)$ equals zero, which occurs when $x = \frac{\ln 7}{\ln 7}$ (since $\ln 1 = 0$). So, $x = 1$ is a critical point.

Step 2: Determine the sign of each factor for different intervals of $x$ Now, we will check the sign of each factor for different intervals of $x$:

1. When $x < 1$:

   • $5^x$ is positive for all $x$.
   • $(13^x + 2)$ is positive for all $x$.
   • $(7^x - 1)$ is negative for all $x$ since $7^x$ is always greater than 1.

   Therefore, $(5^x - 5)$ is positive and $(7^x - 1)$ is negative. The overall sign is negative.

2. When $x = 1$: We already identified this as a critical point, but the inequality is not defined at this point since it would lead to division by zero.

3. When $x > 1$:

   • $5^x$ is positive for all $x$.
   • $(13^x + 2)$ is positive for all $x$.
   • $(7^x - 1)$ is positive for all $x$ greater than 1.

   Therefore, $(5^x - 5)$ is positive, and $(7^x - 1)$ is also positive. The overall sign is positive.

Step 3: Final solution From the analysis above, the inequality is satisfied for $x < 1$ and $x > 1$, except for $x = 1$ (where it is undefined). So, the solution to the inequality is:

$x \in (-\infty, 1) \cup (1, \infty)$

In interval notation, the solution is $(- \infty, 1) \cup (1, \infty)$.