Question

(X^3-3x^2-x+3)\(x^2+3x+2)>0

Answer 1

Надо разложить, перемножить.
Вот ответ: х^2-4х+3/х+2

Answer 2

To solve the inequality $\frac{{x^3 - 3x^2 - x + 3}}{{x^2 + 3x + 2}} > 0$, you can follow these steps:

1. Factor both the numerator and denominator: $\frac{{(x - 1)(x - 1)(x + 3)}}{{(x + 1)(x + 2)}} > 0$

2. Find the critical values by setting both the numerator and denominator equal to zero and solving for $x$: Numerator: $x - 1 = 0 \Rightarrow x = 1$ Denominator: $x + 1 = 0 \Rightarrow x = -1$ and $x + 2 = 0 \Rightarrow x = -2$

3. Use these critical values to create test intervals. You can consider four intervals: $(-∞, -2)$, $(-2, -1)$, $(-1, 1)$, and $(1, ∞)$.

4. Choose a test point in each interval and evaluate the expression $\frac{{(x - 1)(x - 1)(x + 3)}}{{(x + 1)(x + 2)}}$ at that point to determine whether it's greater than 0 or not.

   • For the interval $(-∞, -2)$, choose $x = -3$. You get: $\frac{{(-3 - 1)(-3 - 1)(-3 + 3)}}{{(-3 + 1)(-3 + 2)}} = \frac{{-16}}{{12}} < 0$

   • For the interval $(-2, -1)$, choose $x = -1.5$. You get: $\frac{{(-1.5 - 1)(-1.5 - 1)(-1.5 + 3)}}{{(-1.5 + 1)(-1.5 + 2)}} = \frac{{-0.125}}{{0.75}} < 0$

   • For the interval $(-1, 1)$, choose $x = 0$. You get: $\frac{{(0 - 1)(0 - 1)(0 + 3)}}{{(0 + 1)(0 + 2)}} = \frac{{0}}{{2}} = 0$

   • For the interval $(1, ∞)$, choose $x = 2$. You get: $\frac{{(2 - 1)(2 - 1)(2 + 3)}}{{(2 + 1)(2 + 2)}} = \frac{{5}}{{15}} > 0$

5. Based on the results, you can see that the expression is greater than 0 in the intervals $(-1, 1)$ and $(1, ∞)$.

So, the solution to the inequality $\frac{{x^3 - 3x^2 - x + 3}}{{x^2 + 3x + 2}} > 0$ is $x \in (-1, 1) \cup (1, ∞)$.