Question

((x+3)*(2x+3)):((2-3x)*(2-x))>0

Answer 1

(2x^2+3x+6x+9)(4-2x-6x+3x^2) - после сокращения получается
(2x+1+1+9)/(4-1-1+3x)=2x+11/2+3x   =>
2x+11=2+3x
-x=-9
x=9
9>0 
Ответ: 9

Answer 2

To solve the inequality $((x+3)(2x+3))/((2-3x)(2-x)) > 0$, we can follow these steps:

Step 1: Determine the critical points by setting each factor equal to zero and finding the solutions.

Setting $x + 3 = 0$, we find $x = -3$. Setting $2x + 3 = 0$, we find $x = -\frac{3}{2}$. Setting $2 - 3x = 0$, we find $x = \frac{2}{3}$. Setting $2 - x = 0$, we find $x = 2$.

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

On the number line, we have: $-\infty \quad -3 \quad -\frac{3}{2} \quad \frac{2}{3} \quad 2 \quad \infty$

Step 3: Test the inequality in each interval formed by the critical points.

In the interval $-\infty < x < -3$, we choose a test point $x = -4$. Plugging it into the inequality, we have: $((x+3)(2x+3))/((2-3x)(2-x)) = ((-4+3)(2(-4)+3))/((2-3(-4))(2-(-4))) = \frac{-1}{6} > 0$

In the interval $-3 < x < -\frac{3}{2}$, we choose a test point $x = -2$. Plugging it into the inequality, we have: $((x+3)(2x+3))/((2-3x)(2-x)) = ((-2+3)(2(-2)+3))/((2-3(-2))(2-(-2))) = \frac{5}{8} > 0$

In the interval $-\frac{3}{2} < x < \frac{2}{3}$, we choose a test point $x = 0$. Plugging it into the inequality, we have: $((x+3)(2x+3))/((2-3x)(2-x)) = ((0+3)(2(0)+3))/((2-3(0))(2-(0))) = \frac{9}{4} > 0$

In the interval $\frac{2}{3} < x < 2$, we choose a test point $x = 1$. Plugging it into the inequality, we have: $((x+3)(2x+3))/((2-3x)(2-x)) = ((1+3)(2(1)+3))/((2-3(1))(2-(1))) = \frac{20}{5} > 0$

In the interval $2 < x < \infty$, we choose a test point $x = 3$. Plugging it into the inequality, we have: $((x+3)(2x+3))/((2-3x)(2-x)) = ((3+3)(2(3)+3))/((2-3(3))(2-(3))) = \frac{-36}{35} < 0$

Step 4: Determine the solution.

The solution to the inequality $((x+3)(2x+3))/((2-3x)(2-x)) > 0$