Let solution of inequality `(16^((1)/(x)))/2^(x+3)gt1` is `(-oo, a) cup (b, c)` then

Options

A. `a^(2) + b^(2) + 9c^(2) = 25`
B. `a + b + c gt 0`
C. `a^(2) + b^(2) + c^(2) = 17`
D. `a + b + c lt 0`

Explanation

To solve the inequality \(\frac{16^{\frac{1}{x}}}{2^{x+3}} > 1\), we will follow these steps:

### Step 1: Rewrite the inequality
We can express \(16\) as \(2^4\):
\[ \frac{(2^4)^{\frac{1}{x}}}{2^{x+3}} > 1 \]
This simplifies to:
\[ \frac{2^{\frac{4}{x}}}{2^{x+3}} > 1 \]

### Step 2: Simplify the expression
Using the property of exponents \(\frac{a^m}{a^n} = a^{m-n}\), we can rewrite the inequality as:
\[ 2^{\frac{4}{x} - (x + 3)} > 1 \]
Since \(1\) can be expressed as \(2^0\), we have:
\[ \frac{4}{x} - (x + 3) > 0 \]

### Step 3: Combine the terms
Rearranging the inequality gives:
\[ \frac{4}{x} - x - 3 > 0 \]
To combine the terms, we will take a common denominator \(x\):
\[ \frac{4 - x^2 - 3x}{x} > 0 \]
This simplifies to:
\[ \frac{-x^2 - 3x + 4}{x} > 0 \]

### Step 4: Factor the numerator
We can factor the quadratic expression in the numerator:
\[ -x^2 - 3x + 4 = -(x^2 + 3x - 4) \]
Factoring \(x^2 + 3x - 4\) gives:
\[ x^2 + 3x - 4 = (x + 4)(x - 1) \]
Thus, we have:
\[ \frac{-(x + 4)(x - 1)}{x} > 0 \]

### Step 5: Determine the critical points
The critical points from the factors are:
- \(x = -4\)
- \(x = 1\)
- \(x = 0\) (from the denominator)

### Step 6: Analyze the sign of the expression
We will test the intervals determined by the critical points: \((-∞, -4)\), \((-4, 0)\), \((0, 1)\), and \((1, ∞)\).

1. **Interval \((-∞, -4)\)**: Choose \(x = -5\)
\[ \frac{-(-5 + 4)(-5 - 1)}{-5} = \frac{-(-1)(-6)}{-5} = \frac{-6}{-5} > 0 \]
(Positive)

2. **Interval \((-4, 0)\)**: Choose \(x = -1\)
\[ \frac{-(-1 + 4)(-1 - 1)}{-1} = \frac{-3 \cdot -2}{-1} = \frac{6}{-1} < 0 \]
(Negative)

3. **Interval \((0, 1)\)**: Choose \(x = \frac{1}{2}\)
\[ \frac{-\left(\frac{1}{2} + 4\right)\left(\frac{1}{2} - 1\right)}{\frac{1}{2}} = \frac{-\left(\frac{9}{2}\right)\left(-\frac{1}{2}\right)}{\frac{1}{2}} = \frac{9}{2} > 0 \]
(Positive)

4. **Interval \((1, ∞)\)**: Choose \(x = 2\)
\[ \frac{-(2 + 4)(2 - 1)}{2} = \frac{-6 \cdot 1}{2} < 0 \]
(Negative)

### Step 7: Write the solution
The intervals where the expression is positive are:
\[ (-\infty, -4) \cup (0, 1) \]
Thus, the solution to the inequality is:
\[ (-\infty, -4) \cup (0, 1) \]

### Final Result
The values of \(a\), \(b\), and \(c\) are:
- \(a = -4\)
- \(b = 0\)
- \(c = 1\)