Question

X^3 + 2x^2 - 3x / (x - 1)(x - 2) =< 0

Answer 1

$\frac{x^{3}+2x^{2} -3x }{(x-1)(x-2)}\leq0\\\\\frac{x(x^{2}+2x-3)}{(x-1)(x-2)}\leq0\\\\\frac{x(x-1)(x+3)}{(x-1)(x-2)}\leq0\\\\x(x-1)^{2}(x-2)(x+3)\leq 0,x\neq 1;x\neq2$

    -               +            -              -             +

_____[-3]_____[0]_____(1)_____(2)_____ x

////////////               ////////////  //////////////

Ответ : x ∈ (- ∞ ; - 3] ∪ [0 ; 1) ∪ (1 ; 2)

Answer 2

Объяснение:

$></p> <p>ОДЗ: х-1≠0   х≠1    х-2≠0   </p> <p><img src=$

-∞__-__-3__+__0_-_(1)_-_2__+__+∞

Ответ: х∈(-∞;-3]U[0;1)U(1;2).

Answer 3

To solve the inequality $\frac{x^3 + 2x^2 - 3x}{(x - 1)(x - 2)} \leq 0$, you can follow these steps:

1. Find the critical points by setting the numerator and denominator equal to zero and solving for $x$. The critical points are the values of $x$ that make the expression equal to zero or undefined.

   Numerator: $x^3 + 2x^2 - 3x = 0$

   Factor the numerator:

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

   Factor the quadratic expression in parentheses:

   $x(x + 3)(x - 1) = 0$

   So, the critical points are $x = 0$, $x = -3$, and $x = 1$.

   Denominator: $x - 1 = 0$ and $x - 2 = 0$

   So, the critical points are $x = 1$ and $x = 2$.

2. Create a sign chart to determine the sign of the expression in each interval between the critical points. You can choose test points in each interval to do this.

   The critical points are $x = -3$, $x = 0$, $x = 1$, and $x = 2$.

   Test $x = -4$ in the interval $(-∞, -3)$:

   $\frac{(-4)^3 + 2(-4)^2 - 3(-4)}{(-4 - 1)(-4 - 2)} = \frac{-64 + 32 + 12}{(-5)(-6)} = \frac{-20}{30} = -\frac{2}{3}$

   Test $x = -2$ in the interval $(-3, 0)$:

   $\frac{(-2)^3 + 2(-2)^2 - 3(-2)}{(-2 - 1)(-2 - 2)} = \frac{-8 + 8 + 6}{(-3)(-4)} = \frac{6}{12} = \frac{1}{2}$

   Test $x = 0.5$ in the interval $(0, 1)$:

   $\frac{(0.5)^3 + 2(0.5)^2 - 3(0.5)}{(0.5 - 1)(0.5 - 2)} = \frac{0.125 + 0.25 - 1.5}{(-0.5)(-1.5)} = \frac{-1.125}{0.75} = -1.5$

   Test $x = 1.5$ in the interval $(1, 2)$:

   $\frac{(1.5)^3 + 2(1.5)^2 - 3(1.5)}{(1.5 - 1)(1.5 - 2)} = \frac{3.375 + 4.5 - 4.5}{(0.5)(-0.5)} = \frac{3.375}{-0.25} = -13.5$

   Test $x = 3$ in the interval $(2, ∞)$:

   $\frac{3^3 + 2(3)^2 - 3(3)}{(3 - 1)(3 - 2)} = \frac{27 + 18 - 9}{(2)(1)} = \frac{36}{2} = 18$

3. Now, based on the sign chart, determine when the expression is less than or equal to zero. It is less than or equal to zero when it is negative (less than zero) or zero itself.

   • The expression is less than zero in the intervals $(-∞, -3)$, $(0, 1)$, and $(2, ∞)$