Question

Решение неравенство x^2-64/x>0

Answer 1

$\bf x^2-\dfrac{64}{x} >0\\ \dfrac{x^3-64}{x}>0\\ \dfrac{(x-4)(x^2+4x+16)}{x}>0$

"+" "-" "+"

_(0)_(4)_

Ответ: x∈(-∞; 0)U(4; +∞)

Answer 2

To solve the inequality $\frac{x^2 - 64}{x} > 0$, we need to find the intervals for which the expression is greater than zero. To do this, follow these steps:

Step 1: Factor the expression. Step 2: Determine the critical points (where the expression becomes zero or undefined). Step 3: Create a sign chart to find the intervals where the expression is positive. Step 4: Express the solution as an inequality.

Let's proceed with the steps:

Step 1: Factor the expression. The expression $\frac{x^2 - 64}{x}$ can be factored as $\frac{(x+8)(x-8)}{x}$.

Step 2: Determine the critical points (where the expression becomes zero or undefined). The critical points occur when the numerator is equal to zero (i.e., $(x+8)(x-8) = 0$) or when the denominator is zero (i.e., $x = 0$).

So the critical points are $x = -8, x = 8, x = 0$.

Step 3: Create a sign chart to find the intervals where the expression is positive.

We'll create a sign chart with the intervals separated by the critical points (-8, 0, 8):

$$\begin{array}{|c|c|c|c|c|c|} \hline \text{Interval} & (-\infty, -8) & (-8, 0) & (0, 8) & (8, \infty) \\ \hline x+8 & - & + & + & + \\ \hline x-8 & - & - & + & + \\ \hline x & - & - & + & + \\ \hline \frac{(x+8)(x-8)}{x} & + & - & + & + \\ \hline \end{array}$$

Step 4: Express the solution as an inequality.

From the sign chart, we can see that the expression $\frac{x^2 - 64}{x}$ is positive in two intervals: $(-8, 0)$ and $(8, \infty)$. To express this as an inequality, we combine the two intervals:

$-8 < x < 0 \quad \text{or} \quad x > 8$

Thus, the solution to the inequality $\frac{x^2 - 64}{x} > 0$ is $x \in (-8, 0) \cup (8, \infty)$.