Question

(x-2)(x+3)(x-5)/2x+1 больше либо равно 0

Answer 1

(-∞;-3]∪(-0.5;2]∪[5;∞)

Answer 2

To determine the sign of the expression $\frac{(x-2)(x+3)(x-5)}{2x+1}$, we need to identify the intervals where the expression is greater than or equal to zero.

To do this, we can use the concept of interval notation and the properties of the expression:

1. Set the numerator of the fraction $(x-2)(x+3)(x-5)$ equal to zero and find the critical points where the numerator could change sign.
2. Find the critical points of the denominator $2x+1$ and avoid values that make the denominator zero since division by zero is undefined.
3. Check the sign of the expression in the intervals formed by the critical points.

Let's go through the steps:

1. Set the numerator equal to zero and solve for $x$:

$(x-2)(x+3)(x-5) = 0$

The critical points are $x = 2$, $x = -3$, and $x = 5$.

2. Find the critical points of the denominator $2x+1$:

$2x + 1 = 0$

$2x = -1$

$x = -\frac{1}{2}$

3. Now, we have five intervals to check:

• Interval 1: $x < -3$
• Interval 2: $-3 < x < -\frac{1}{2}$
• Interval 3: $-\frac{1}{2} < x < 2$
• Interval 4: $2 < x < 5$
• Interval 5: $x > 5$

Test a value from each interval in the expression to determine the sign:

• Interval 1 (e.g., $x = -4$):

$\frac{(-4-2)(-4+3)(-4-5)}{2(-4)+1} = \frac{(-6)(-1)(-9)}{-7} > 0$

• Interval 2 (e.g., $x = -1$):

$\frac{(-1-2)(-1+3)(-1-5)}{2(-1)+1} = \frac{(-3)(2)(-6)}{-1} > 0$

• Interval 3 (e.g., $x = 0$):

$\frac{(0-2)(0+3)(0-5)}{2(0)+1} = \frac{(-2)(3)(-5)}{1} < 0$

• Interval 4 (e.g., $x = 3$):

$\frac{(3-2)(3+3)(3-5)}{2(3)+1} = \frac{(1)(6)(-2)}{7} < 0$

• Interval 5 (e.g., $x = 6$):

$\frac{(6-2)(6+3)(6-5)}{2(6)+1} = \frac{(4)(9)(1)}{13} > 0$

Therefore, the expression $\frac{(x-2)(x+3)(x-5)}{2x+1}$ is greater than or equal to zero in the intervals $(-3, -\frac{1}{2})$ and $(5, \infty)$.