Solve `((2x+3)(4-3x)^3(x-4))/((x-2)^2x^5)le0 `

Explanation

To solve the inequality

\[ \frac{(2x + 3)(4 - 3x)^3(x - 4)}{(x - 2)^2 x^5} \leq 0, \]

we will follow these steps:

### Step 1: Identify the critical points
We need to find the values of \(x\) that make the numerator and denominator equal to zero.

**Numerator:**
1. \(2x + 3 = 0 \Rightarrow x = -\frac{3}{2}\)
2. \(4 - 3x = 0 \Rightarrow x = \frac{4}{3}\) (this term is cubed, so it will affect the sign of the function)
3. \(x - 4 = 0 \Rightarrow x = 4\)

**Denominator:**
1. \((x - 2)^2 = 0 \Rightarrow x = 2\) (this term is squared, so it will not change the sign)
2. \(x^5 = 0 \Rightarrow x = 0\)

**Critical Points:**
The critical points are \(x = -\frac{3}{2}, 0, \frac{4}{3}, 2, 4\).

### Step 2: Create a number line
Place the critical points on a number line:

\[ -\frac{3}{2}, 0, \frac{4}{3}, 2, 4 \]

### Step 3: Test intervals
We will test the sign of the expression in the intervals defined by these critical points:

1. **Interval \((- \infty, -\frac{3}{2})\)**: Choose \(x = -2\)
\[ \frac{(2(-2) + 3)(4 - 3(-2))^3(-2 - 4)}{((-2 - 2)^2)(-2)^5} = \frac{(-4 + 3)(4 + 6)^3(-6)}{(4)(-32)} = \frac{(-1)(10^3)(-6)}{(-128)} > 0 \]

2. **Interval \((- \frac{3}{2}, 0)\)**: Choose \(x = -1\)
\[ \frac{(2(-1) + 3)(4 - 3(-1))^3(-1 - 4)}{((-1 - 2)^2)(-1)^5} = \frac{(1)(7^3)(-5)}{(9)(-1)} < 0 \]

3. **Interval \((0, \frac{4}{3})\)**: Choose \(x = 1\)
\[ \frac{(2(1) + 3)(4 - 3(1))^3(1 - 4)}{((1 - 2)^2)(1)^5} = \frac{(5)(1^3)(-3)}{(1)(1)} < 0 \]

4. **Interval \((\frac{4}{3}, 2)\)**: Choose \(x = 1.5\)
\[ \frac{(2(1.5) + 3)(4 - 3(1.5))^3(1.5 - 4)}{((1.5 - 2)^2)(1.5)^5} = \frac{(6)(-0.5^3)(-2.5)}{(0.25)(7.59375)} > 0 \]

5. **Interval \((2, 4)\)**: Choose \(x = 3\)
\[ \frac{(2(3) + 3)(4 - 3(3))^3(3 - 4)}{((3 - 2)^2)(3)^5} = \frac{(9)(-5^3)(-1)}{(1)(243)} < 0 \]

6. **Interval \((4, \infty)\)**: Choose \(x = 5\)
\[ \frac{(2(5) + 3)(4 - 3(5))^3(5 - 4)}{((5 - 2)^2)(5)^5} = \frac{(13)(-11^3)(1)}{(9)(3125)} < 0 \]

### Step 4: Determine the sign of the expression
From the tests, we have:
- Positive in \((- \infty, -\frac{3}{2})\)
- Negative in \((- \frac{3}{2}, 0)\)
- Negative in \((0, \frac{4}{3})\)
- Positive in \((\frac{4}{3}, 2)\)
- Negative in \((2, 4)\)
- Negative in \((4, \infty)\)

### Step 5: Write the solution
We need the intervals where the expression is less than or equal to zero, including the points where the numerator is zero but excluding the points where the denominator is zero:

\[ x \in \left(-\infty, -\frac{3}{2}\right) \cup \left(-\frac{3}{2}, 0\right) \cup (0, \frac{4}{3}] \cup (2, 4) \]

### Final Answer
The solution to the inequality is:

\[ x \in \left(-\infty, -\frac{3}{2}\right) \cup \left(-\frac{3}{2}, 0\right) \cup \left(0, \frac{4}{3}\right] \cup (2, 4) \]