Question

X-5/2x-x^2 больше либо равно 0

Answer 1

Сначала мы всё умножаем на знаменатель, чтобы от него избавиться. Следовательно 0 становится знаменателем. Потом мы переносим в лесу сторону всё с Х, а в правую сторону без него

Answer 2

To determine when the expression $\frac{X-5}{2x-x^2}$ is greater than or equal to 0, we need to find the critical points and the intervals where the expression is positive or zero.

Step 1: Find the critical points. Critical points occur when the expression is either equal to zero or undefined (denominator equals zero). So, we need to find when the denominator $2x-x^2$ is equal to zero:

$2x-x^2 = 0$

Step 2: Solve for x.

$x^2 - 2x = 0$

Factor the equation:

$x(x - 2) = 0$

Setting each factor to zero:

$x = 0$ or $x - 2 = 0$

So, the critical points are $x = 0$ and $x = 2$.

Step 3: Create intervals and test the expression.

We can create three intervals: $-\infty < x < 0$, $0 < x < 2$, and $2 < x < \infty$. We will test the expression $\frac{X-5}{2x-x^2}$ in each interval to determine its sign.

For $x < 0$: Choose $x = -1$ (a value less than 0): $\frac{-1-5}{2(-1)-(-1)^2} = \frac{-6}{-2+1} = -6$, which is negative.

For $0 < x < 2$: Choose $x = 1$ (a value between 0 and 2): $\frac{1-5}{2(1)-1^2} = \frac{-4}{2-1} = -4$, which is negative.

For $x > 2$: Choose $x = 3$ (a value greater than 2): $\frac{3-5}{2(3)-3^2} = \frac{-2}{6-9} = \frac{-2}{-3} = \frac{2}{3}$, which is positive.

Step 4: Conclusion The expression $\frac{X-5}{2x-x^2}$ is greater than or equal to 0 for $x > 2$. In the intervals $-\infty < x < 0$ and $0 < x < 2$, the expression is negative.

So, the solution to the inequality $\frac{X-5}{2x-x^2} \geq 0$ is $x > 2$.