x-1/7-5x>0

To solve the inequality \( \frac{x-1}{7-5x} > 0 \), we need to determine where the expression is positive. This occurs when both the numerator and denominator are either both positive or both negative.

Step 1: Find where the numerator \( x - 1 > 0 \)

\[ x - 1 > 0 \implies x > 1 \]

Step 2: Find where the denominator \( 7 - 5x > 0 \)

\[ 7 - 5x > 0 \implies 7 > 5x \implies x < \frac{7}{5} \]

Step 3: Determine intervals

Now we have two conditions:

1. \( x > 1 \)
2. \( x < \frac{7}{5} \)

Combining these conditions, we find:

\[ 1 < x < \frac{7}{5} \]

Step 4: Sign analysis of critical points

Next, we need to consider the critical points:

• The numerator changes sign at \( x = 1 \)
• The denominator changes sign at \( x = \frac{7}{5} \)

We can summarize the sign chart:

1. For \( x < 1 \): \( x - 1 < 0 \) and \( 7 - 5x > 0 \) → Overall negative.
2. For \( 1 < x < \frac{7}{5} \): \( x - 1 > 0 \) and \( 7 - 5x > 0 \) → Overall positive.
3. For \( x > \frac{7}{5} \): \( x - 1 > 0 \) and \( 7 - 5x < 0 \) → Overall negative.

Final Solution: The solution to the inequality \( \frac{x-1}{7-5x} > 0 \) is:

\[ x \in (1, \frac{7}{5}) \]

So, the final answer is:

\[ \boxed{(1, \frac{7}{5})} \]