If `((x-3)^((-|x|)/x) sqrt((x-4)^(2))(17-x))/(sqrt(-x)(-x^(2)+x-1)(|x|-32))lt0` then no. of integers `x` satisfying the inequality is:

Explanation

To solve the inequality

\[ \frac{(x-3)^{-\frac{|x|}{x}} \sqrt{(x-4)^{2}}(17-x)}{\sqrt{-x}(-x^{2}+x-1)(|x|-32)} < 0, \]

we will analyze each component step by step.

### Step 1: Analyze the components of the inequality

1. **Numerator:**
- \( (x-3)^{-\frac{|x|}{x}} \)
- \( \sqrt{(x-4)^{2}} = |x-4| \)
- \( (17-x) \)

2. **Denominator:**
- \( \sqrt{-x} \)
- \( -x^{2}+x-1 \)
- \( |x|-32 \)

### Step 2: Determine the conditions for the numerator and denominator

**Numerator:**
- \( (x-3)^{-\frac{|x|}{x}} \) is defined for \( x \neq 3 \) and is positive if \( x > 3 \) or \( x < 0 \) (since \(-\frac{|x|}{x}\) will be negative for positive \(x\)).
- \( |x-4| \) is positive for \( x \neq 4 \).
- \( (17-x) \) is positive for \( x < 17 \).

**Denominator:**
- \( \sqrt{-x} \) is defined for \( x < 0 \).
- \( -x^{2}+x-1 = -(x^{2}-x+1) \) is always negative since the discriminant \(1 - 4 < 0\).
- \( |x|-32 \) is negative for \( -32 < x < 32 \).

### Step 3: Combine the conditions

1. **From the numerator:**
- \( x < 17 \) and \( x \neq 3 \), \( x \neq 4 \).

2. **From the denominator:**
- \( x < 0 \) (from \(\sqrt{-x}\)).
- \( |x|-32 < 0 \) implies \( -32 < x < 32 \) (but we are only considering \(x < 0\)).

### Step 4: Find the valid range for \(x\)

From the conditions, we have:
- \( x < 0 \)
- \( x < 17 \)
- \( x \neq 3 \)
- \( x \neq 4 \)

The most restrictive condition is \( x < 0 \). Thus, we need to find integers in the range \( -32 < x < 0 \).

### Step 5: Count the integers

The integers satisfying \( -32 < x < 0 \) are:
\[ -31, -30, -29, \ldots, -1. \]
This gives us a total of \(31\) integers.

### Final Answer

Thus, the number of integers \( x \) satisfying the inequality is:

\[ \boxed{31}. \]