Question

Неравенство 5^2/х≤0.2^х-3

Answer 1

$5^{\frac{2}{x}}=(0,2)^{x-3}\; ,\; \; \; ODZ:\; \; x\ne 0\\\\0,2=\frac{2}{10}=\frac{1}{5}=5^{-1}\\\\5^{\frac{2}{x}}=5^{-x+3}\\\\ \frac{2}{x}=-x+3\\\\2=-x^2+3x\; ,\; \; x^2-3x+2=0\; ,\\\\\underline {x_1=1\; ,\; x_2=2}$

Answer 2

To solve the inequality $\frac{5^2}{x} \leq 0.2^{x-3}$, we need to find the possible values of $x$ that satisfy the inequality.

First, let's simplify the inequality by expressing both sides with the same base. We know that $5^2 = 25$ and $0.2 = \frac{1}{5}$, so:

$\frac{5^2}{x} \leq 0.2^{x-3}$

$\frac{25}{x} \leq \left(\frac{1}{5}\right)^{x-3}$

Now, we can rewrite $\left(\frac{1}{5}\right)^{x-3}$ as $\frac{1}{5^{x-3}}$:

$\frac{25}{x} \leq \frac{1}{5^{x-3}}$

Next, we can get rid of the fractions by multiplying both sides of the inequality by $5^x$ (note that $5^x$ is always positive for any real $x$):

$25 \cdot 5^x \leq x$

Now, we need to consider two cases:

Case 1: $x > 0$

In this case, $x$ is positive, and we can safely divide both sides of the inequality by $x$ without changing the inequality's direction:

$25 \cdot 5^x/x \leq x/x$

$25 \cdot 5^x \leq 1$

However, there is no real solution for this inequality because the left-hand side ($25 \cdot 5^x$) is always greater than 1 for any positive real $x$.

Case 2: $x < 0$

In this case, $x$ is negative, and when we divide both sides by $x$, we need to reverse the inequality's direction:

$25 \cdot 5^x/x \geq x/x$

$25 \cdot 5^x \geq 1$

Now, this inequality has a solution. Since $5^x$ becomes arbitrarily large as $x$ becomes more negative, we can find a value of $x$ that satisfies this inequality. Therefore, the solution for $x$ is $x < 0$.

To summarize:

The solution for the inequality $\frac{5^2}{x} \leq 0.2^{x-3}$ is $x < 0$.