√(3x - 6)/(1 - x)+ 5 > 0

To solve the inequality \(\sqrt{\frac{3x - 6}{1 - x}} + 5 > 0\), we can start by isolating the square root term.

Step 1: Isolate the Square Root

Subtract 5 from both sides:

\[ \sqrt{\frac{3x - 6}{1 - x}} > -5 \]

Since the square root is always non-negative, this inequality is always true. Thus, we need to ensure that the expression inside the square root is defined and non-negative.

Step 2: Set the Inequality Inside the Square Root

We need to solve:

\[ \frac{3x - 6}{1 - x} \geq 0 \]

Step 3: Find Critical Points

The critical points occur where the numerator and denominator are zero:

1. Numerator: \(3x - 6 = 0\) \[ 3x = 6 \implies x = 2 \]

2. Denominator: \(1 - x = 0\) \[ x = 1 \]

Step 4: Test Intervals

We will analyze the sign of the expression \(\frac{3x - 6}{1 - x}\) in the intervals determined by the critical points \(x = 1\) and \(x = 2\).

• Intervals: \((-\infty, 1)\), \((1, 2)\), \((2, \infty)\)

1. Interval \((-\infty, 1)\): Choose \(x = 0\) \[ \frac{3(0) - 6}{1 - 0} = \frac{-6}{1} < 0 \quad \text{(negative)} \]

2. Interval \((1, 2)\): Choose \(x = 1.5\) \[ \frac{3(1.5) - 6}{1 - 1.5} = \frac{4.5 - 6}{1 - 1.5} = \frac{-1.5}{-0.5} = 3 \quad \text{(positive)} \]

3. Interval \((2, \infty)\): Choose \(x = 3\) \[ \frac{3(3) - 6}{1 - 3} = \frac{9 - 6}{1 - 3} = \frac{3}{-2} < 0 \quad \text{(negative)} \]

Step 5: Determine the Valid Intervals

From the tests, we find that \(\frac{3x - 6}{1 - x} \geq 0\) in the interval \((1, 2]\).

Step 6: Conclusion

Thus, the solution to the inequality \(\sqrt{\frac{3x - 6}{1 - x}} + 5 > 0\) is:

\[ x \in (1, 2] \]

This means \(x\) can take any value greater than 1 and less than or equal to 2.