Question

X(x^2-9) (5x-6-x^2)<=0

Answer 1

Ответ:

Объяснение:

x(x^2-9) (5x-6-x^2)<=0

x(x-3)(x+3)(-x^2+5x-6)<=0

(-x^2+5x-6)x(x-3)(x+3)<=0

-(x²-5x+6)x(x-3)(x+3)<=0

(x²-5x+6)x(x-3)(x+3)≥0

(x²-5x+6)x(x-3)(x+3)=0

x²-5x+6=0

D=25-24=1

√1=1

x=(5±1)/2 = 3 ; 2

x = 0

x-3=0

x=3

x+3=0

x= -3

x∈[-3;0] ∪ [2;+∞)

Answer 2

To solve the inequality $X(x^2-9)(5x-6-x^2) \leq 0$, you can follow these steps:

1. First, find the critical points of the expression by setting each factor equal to zero and solving for $x$:

   • $x^2 - 9 = 0$
   • $5x - 6 - x^2 = 0$

Let's solve these equations:

For $x^2 - 9 = 0$, you get: $x^2 = 9$ $x = \pm 3$

For $5x - 6 - x^2 = 0$, you get: $x^2 - 5x + 6 = 0$ $(x - 2)(x - 3) = 0$ So, $x = 2$ or $x = 3$.

2. Now, you have four critical points: $x = -3$, $x = 2$, $x = 3$, and $x = -3$. These divide the real number line into five intervals:

   • $(-∞, -3)$
   • $(-3, 2)$
   • $(2, 3)$
   • $(3, ∞)$

3. Pick a test point from each interval and evaluate the expression $X(x^2-9)(5x-6-x^2)$ to determine its sign in that interval.

   • For the interval $(-∞, -3)$, pick a test point like $x = -4$: $X(-4^2 - 9)(5(-4) - 6 - (-4)^2) = X(-16 - 9)(-20 - 16) = X(-25)(-36) = 900X$

   • For the interval $(-3, 2)$, pick a test point like $x = 0$: $X(0^2 - 9)(5(0) - 6 - (0)^2) = X(-9)(-6) = 54X$

   • For the interval $(2, 3)$, pick a test point like $x = 2.5$: $X(2.5^2 - 9)(5(2.5) - 6 - (2.5)^2) = X(2.25)(6.5 - 6 - 6.25) = X(2.25)(-0.75) = -1.6875X$

   • For the interval $(3, ∞)$, pick a test point like $x = 4$: $X(4^2 - 9)(5(4) - 6 - (4)^2) = X(7)(20 - 6 - 16) = X(7)(-2) = -14X$

4. Now, consider the sign of $X(x^2-9)(5x-6-x^2)$ in each interval:

   • In $(-∞, -3)$, the expression is positive or zero when $X \geq 0$.
   • In $(-3, 2)$, the expression is positive or zero when $X \leq 0$.
   • In $(2, 3)$, the expression is negative when $X \geq 0$.
   • In $(3, ∞)$, the expression is negative when $X \leq 0$.

5. Finally, combine the intervals where the expression is non-negative and non-positive:

   • The solution to $X(x^2-9)(5x-6-x^2) \leq 0$ is: $X \geq 0 \text{ for } x \in (-∞, -3) \cup (2, 3)$ $X \leq 0 \text{ for } x \in (-3, 2) \cup (3, ∞)$