Question

X^2+2x/x+3≤0 решите неравенство

Answer 1

X^2+x+3≤
D=1-12
D<0
решений нет

Answer 2

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

Step 1: Find the critical points by setting the expression inside the fraction equal to zero and solving for $x$. Step 2: Determine the sign of the expression in each interval formed by the critical points. Step 3: Identify the intervals where the expression is less than or equal to zero.

Let's start with Step 1:

$\frac{x^2+2x}{x+3} = 0$

The expression inside the fraction is a quadratic, so we can find the critical points by solving the quadratic equation:

$x^2 + 2x = 0$

Factoring out $x$:

$x(x + 2) = 0$

Setting each factor to zero:

1. $x = 0$
2. $x + 2 = 0$ $x = -2$

Now, we have the critical points $x = 0$ and $x = -2$. These points divide the number line into three intervals:

1. $x < -2$
2. $-2 < x < 0$
3. $x > 0$

Next, let's determine the sign of the expression $\frac{x^2+2x}{x+3}$ in each interval.

For $x < -2$: Choose $x = -3$ (a value less than -2): $\frac{(-3)^2+2(-3)}{-3+3} = \frac{9-6}{0} = \frac{3}{0}$ (Note: Division by zero is not defined.)

For $-2 < x < 0$: Choose $x = -1$ (a value between -2 and 0): $\frac{(-1)^2+2(-1)}{-1+3} = \frac{1-2}{2} = \frac{-1}{2}$

For $x > 0$: Choose $x = 1$ (a value greater than 0): $\frac{1^2+2(1)}{1+3} = \frac{1+2}{4} = \frac{3}{4}$

Now, we have the signs of the expression in each interval:

1. $x < -2$: undefined (division by zero)
2. $-2 < x < 0$: $\frac{x^2+2x}{x+3} < 0$ (negative)
3. $x > 0$: $\frac{x^2+2x}{x+3} > 0$ (positive)

Finally, we want to find the intervals where the expression is less than or equal to zero ($\leq 0$).

From the analysis above, we see that the expression is less than or equal to zero in the interval $-2 < x < 0$.

So, the solution to the inequality $\frac{x^2+2x}{x+3} \leq 0$ is $-2 < x < 0$.