Question

-xв квадраті(x-1)(x+2)>0

Answer 1

Ответ: (-2;0)U(0;1)

Пошаговое объяснение:

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

Приравняем множители к 0

-x^2=0; x-1=0; x+2=0

Решим

x=0; x=1; x=-2

Проверяем

1) x<-2; 2) -2<x<0; 3) 0<x<1; 4) x>1

Выбираем значение из каждого интервала и подставляем его в начальное неравенство:

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

-36>0 неверно

2) -(-1)^2(-1-1)(-1+2)>0

3>0 верно

3) 0.5^2(0.5-1)(0.5+2)>0

0.3125>0 верно

4) -(2)^2(2-1)(2+2)>0

-16>0 неверно

Решение включает все истинные интервалы. 2) -2<x<0; 3) 0<x<1

Отсюда

(-2;0)U(0;1)

Answer 2

To solve the inequality $-x^2(x-1)(x+2) > 0$, we need to find the intervals for which the expression is greater than zero. To do this, we'll use a sign chart.

Step 1: Find the critical points by setting each factor to zero and solving for $x$: $-x^2 = 0 \Rightarrow x = 0$ (double root), $x - 1 = 0 \Rightarrow x = 1$, $x + 2 = 0 \Rightarrow x = -2$.

Step 2: Plot these critical points on a number line:

pythonCopy code
... -3  -2  -1   0   1   2  ...
    -         +   0   -   +

Step 3: Choose a test point within each interval and evaluate the expression $-x^2(x-1)(x+2)$.

• For $x < -2$, choose $x = -3$ (a value less than -2): $-(-3)^2(-3-1)(-3+2) = -9 \cdot (-4) \cdot (-1) = -36$. The expression is negative.

• For $-2 < x < 0$, choose $x = -1$ (a value between -2 and 0): $-(-1)^2(-1-1)(-1+2) = -1 \cdot (-2) \cdot 1 = 2$. The expression is positive.

• For $0 < x < 1$, choose $x = 0.5$ (a value between 0 and 1): $-(0.5)^2(0.5-1)(0.5+2) = -0.25 \cdot (-0.5) \cdot 2.5 = 0.3125$. The expression is positive.

• For $x > 1$, choose $x = 2$ (a value greater than 1): $-(2)^2(2-1)(2+2) = -4 \cdot 1 \cdot 4 = -16$. The expression is negative.

Step 4: Determine the sign of the expression in each interval and write the solution:

• $-x^2(x-1)(x+2) > 0$ is true when $-2 < x < 0$ and $0 < x < 1$.

So, the solution to the inequality is $-2 < x < 0$ and $0 < x < 1$.