Question

(x-3)(x^2-1) / x^2+3x+2 >0

Answer 1

(x-3)(x²-1) / (x²+3x+2) >0

х²+3х+2=0; х=(-3±√(9-8))/2; х=-1, х=-2

(х-3)(х-1)(х+1)/((х+2)(х+1)) >0

(х-3)(х-1)(х+1)²/((х+2))) >0

решим методом интервалов.

______-2_____-1___1____3___

     -              +         +      -         +

х∈(-2;-1)∪(-1;1)∪(3;+∞)

Answer 2

Ответ:

Пошаговое объяснение:

Решение на фото.

Answer 3

To solve the inequality $\frac{(x-3)(x^2-1)}{x^2+3x+2} > 0$, we need to find the intervals of $x$ for which the expression is greater than zero.

Step 1: Factorize the expressions. $(x-3)$ and $x^2-1$ are already in factored form.

For $x^2+3x+2$, we can find its roots by setting it equal to zero: $x^2 + 3x + 2 = 0$

Using the quadratic formula, we get: $x = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$ where $a = 1$, $b = 3$, and $c = 2$.

$x = \frac{-3 \pm \sqrt{3^2-4 \cdot 1 \cdot 2}}{2 \cdot 1}$ $x = \frac{-3 \pm \sqrt{9-8}}{2}$ $x = \frac{-3 \pm \sqrt{1}}{2}$ $x = \frac{-3 \pm 1}{2}$

So, the roots of $x^2+3x+2$ are $x = -2$ and $x = -1$.

Step 2: Identify the sign of the expression in each interval:

a) $x < -2$: In this interval, all three factors $(x-3)$, $x^2-1$, and $x^2+3x+2$ are negative. Dividing three negative numbers will result in a negative value: $\frac{(x-3)(x^2-1)}{x^2+3x+2} < 0$

b) $-2 < x < -1$: In this interval, $(x-3)$ is negative, $x^2-1$ is positive, and $x^2+3x+2$ is negative. Dividing one negative and two positive numbers will result in a positive value: $\frac{(x-3)(x^2-1)}{x^2+3x+2} > 0$

c) $x > -1$: In this interval, all three factors $(x-3)$, $x^2-1$, and $x^2+3x+2$ are positive. Dividing three positive numbers will result in a positive value: $\frac{(x-3)(x^2-1)}{x^2+3x+2} > 0$

Step 3: Combine the results: The inequality is satisfied when $x$ is in the interval $-2 < x < -1$.

So, the solution to the inequality is $x \in (-2, -1)$.