Question

(х+1)(x+3)/x-2 меньше нуля

Answer 1

(х+1)(x+3)/x-2<0

(х+1)<0 x<-1

(x+3)<0 x<-3

(x-2)<0 x<2

Answer 2

To determine the intervals for which the expression $\frac{(x+1)(x+3)}{x-2}$ is less than zero, we need to find the values of $x$ that make the expression negative.

First, let's find the critical points by setting the expression equal to zero and finding when the denominator is zero (which would make the expression undefined):

1. Set the numerator equal to zero: $(x + 1)(x + 3) = 0$

This gives us two critical points: $x + 1 = 0 \Rightarrow x = -1$ and $x + 3 = 0 \Rightarrow x = -3$

2. Set the denominator equal to zero to find when the expression is undefined: $x - 2 = 0 \Rightarrow x = 2$

Now, we have four intervals on the number line to consider: $(-∞, -3), (-3, -1), (-1, 2), (2, ∞)$. We can test any value within each interval to determine if the expression is negative or positive in that interval:

1. Test $x = -4$ (any value less than -3):
   $\frac{(x+1)(x+3)}{x-2} = \frac{(-4+1)(-4+3)}{-4-2} = \frac{-3}{-6} = \frac{1}{2}$ (positive)

2. Test $x = -2.5$ (any value between -3 and -1):
   $\frac{(x+1)(x+3)}{x-2} = \frac{(-2.5+1)(-2.5+3)}{-2.5-2} = \frac{-1.5}{-4.5} = \frac{1}{3}$ (positive)

3. Test $x = 0$ (any value between -1 and 2):
   $\frac{(x+1)(x+3)}{x-2} = \frac{(0+1)(0+3)}{0-2} = \frac{3}{-2} = -\frac{3}{2}$ (negative)

4. Test $x = 3$ (any value greater than 2):
   $\frac{(x+1)(x+3)}{x-2} = \frac{(3+1)(3+3)}{3-2} = \frac{24}{1} = 24$ (positive)

Now, we can see that the expression is negative in the interval $(-1, 2)$. Thus, the solution to the inequality $\frac{(x+1)(x+3)}{x-2} < 0$ is $x \in (-1, 2)$.