Question

Решите неравенство x²-9/x-5<0

Answer 1

$\frac{(x - 3)(x + 3)}{x - 5} < 0 \\ x = 3 \\ x = - 3 \\ x = 5$
__-_-3__+_3_-__5__+__

(-&;-3)(3;5)

Answer 2

To solve the inequality $\frac{x^2-9}{x-5} < 0$, follow these steps:

Step 1: Find the critical points by setting the numerator and denominator equal to zero and solving for $x$:

$x^2 - 9 = 0$

This is a difference of squares, so we can factor it:

$(x + 3)(x - 3) = 0$

So, $x = -3$ or $x = 3$.

$x - 5 = 0$

This gives $x = 5$.

Step 2: Create an interval chart with these critical points:

Interval	Test Point	$(x+3)(x-3)$	$x-5$	$\frac{(x+3)(x-3)}{x-5}$
$x < -3$	-4	(-1)(-7)	-9	$\frac{7}{9}$
$-3 < x < 3$	0	(-3)(3)	-3	$\frac{-9}{-3} = 3$
$3 < x < 5$	4	(1)(7)	-1	$\frac{-7}{-1} = 7$
$x > 5$	6	(3)(9)	1	$\frac{27}{1} = 27$

Step 3: Determine the sign of $\frac{(x+3)(x-3)}{x-5}$ in each interval.

Step 4: Identify the intervals where $\frac{(x+3)(x-3)}{x-5} < 0$. These are the intervals where the expression is negative.

From the interval chart, we see that the inequality is satisfied for $x \in (-3, 3) \cup (5, \infty)$.

Thus, the solution to the inequality is:

$x \in (-3, 3) \cup (5, \infty)$