Question

(x-1)(x^2+4x+3)/x^2+2>0

Answer 1

думаю что это лучший ответ : )

Answer 2

To solve the inequality $(x-1)(x^2+4x+3)/(x^2+2) > 0$, we need to find the intervals where the expression is greater than zero.

Let's break it down step by step:

1. First, let's find the critical points by setting the numerator and denominator equal to zero and solving for $x$:

Numerator: $(x-1)(x^2+4x+3) = 0$

The factors of the quadratic expression are $(x+1)(x+3)$, so the numerator is zero when $x = 1, -1,$ or $-3$.

Denominator: $x^2+2 = 0$

The denominator has no real solutions since the square of any real number is non-negative, and adding 2 to that will always be positive.

2. Now, we'll create a sign chart to analyze the sign of the expression in different intervals. The critical points and the asymptote (where the denominator approaches zero) will divide the number line into separate intervals.

Number line: (-∞, -3) | (-3, -1) | (-1, 1) | (1, ∞)

3. We'll pick test points from each interval and evaluate the expression to determine its sign.

Test points:

• Choose $x = -4$ (in the interval $(-∞, -3)$)
• Choose $x = -2$ (in the interval $(-3, -1)$)
• Choose $x = 0$ (in the interval $(-1, 1)$)
• Choose $x = 2$ (in the interval $(1, ∞)$)

Evaluate the expression at the test points:

For $x = -4$: $((-4)-1)((-4)^2+4(-4)+3)/( (-4)^2+2) = (-5)(7)/18 < 0$

For $x = -2$: $((-2)-1)((-2)^2+4(-2)+3)/( (-2)^2+2) = (-3)(5)/6 > 0$

For $x = 0$: $(0-1)((0)^2+4(0)+3)/( (0)^2+2) = (-1)(3)/2 < 0$

For $x = 2$: $(2-1)((2)^2+4(2)+3)/( (2)^2+2) = (1)(15)/6 > 0$

4. Finally, we can determine the intervals where the expression is greater than zero:

The expression is greater than zero in the intervals $(-3, -1)$ and $(1, \infty)$.

So, the solution to the inequality is:

$x \in (-3, -1) \cup (1, \infty)$