Question

Решите рационально неравенство (2-3x)(3x+2)(5+3x)(2x-3)>0

Answer 1

Решение задания смотри на фотографии

Answer 2

To solve the rational inequality $(2-3x)(3x+2)(5+3x)(2x-3) > 0$, we need to find the intervals where the expression is greater than zero.

To do this, we can use the concept of intervals and signs. First, let's find the critical points, where the expression becomes zero:

1. Set each factor to zero and solve for $x$:

$(2-3x) = 0 \Rightarrow x = \frac{2}{3}$

$(3x+2) = 0 \Rightarrow x = -\frac{2}{3}$

$(5+3x) = 0 \Rightarrow x = -\frac{5}{3}$

$(2x-3) = 0 \Rightarrow x = \frac{3}{2}$

2. Now, let's mark these critical points on the number line:

pythonCopy code
...  -2  -5/3  -2/3  2/3  3/2  ...

3. Choose test points within each interval:

We'll pick a test point from each interval and evaluate the expression $(2-3x)(3x+2)(5+3x)(2x-3)$ at those points to determine the sign of the expression within that interval.

Let's pick the test points:

• $x = -3$ (a number less than $-2$)
• $x = -1$ (a number between $-5/3$ and $-2/3$)
• $x = 0$ (a number between $-2/3$ and $2/3$)
• $x = 1$ (a number between $2/3$ and $3/2$)
• $x = 2$ (a number greater than $3/2$)

4. Evaluate the expression at the test points:

• $x = -3: (2-3(-3))(3(-3)+2)(5+3(-3))(2(-3)-3) = (11)(-7)(-4)(-9) = 2772$
• $x = -1: (2-3(-1))(3(-1)+2)(5+3(-1))(2(-1)-3) = (5)(-1)(2)(-5) = 50$
• $x = 0: (2-3(0))(3(0)+2)(5+3(0))(2(0)-3) = (2)(2)(5)(-3) = -60$
• $x = 1: (2-3(1))(3(1)+2)(5+3(1))(2(1)-3) = (-1)(5)(8)(-1) = 40$
• $x = 2: (2-3(2))(3(2)+2)(5+3(2))(2(2)-3) = (-4)(8)(11)(1) = -352$

5. Determine the sign of the expression within each interval:

pythonCopy code
...  -2  -5/3  -2/3  2/3  3/2  ...
     +    -    +    -    +

6. Write the solution:

From the sign chart, we see that the expression $(2-3x)(3x+2)(5+3x)(2x-3)$ is greater than zero in two intervals:

1. $-2 < x < -\frac{5}{3}$
2. $\frac{2}{3} < x < \frac{3}{2}$

So, the solution to the rational inequality is:

$-2 < x < -\frac{5}{3}$ or $\frac{2}{3} < x < \frac{3}{2}$