Let (x) is a real function, defines as `f(x) =(x-1)/(x+1),` then prove that `f(2x)=(3f(x)+1)/(f(x)+3).`

Explanation

To prove that \( f(2x) = \frac{3f(x) + 1}{f(x) + 3} \) where \( f(x) = \frac{x-1}{x+1} \), we will follow these steps:

### Step 1: Calculate \( f(2x) \)

We start by substituting \( 2x \) into the function \( f \):

\[ f(2x) = \frac{2x - 1}{2x + 1} \]

### Step 2: Express \( x \) in terms of \( f(x) \)

From the definition of \( f(x) \):

\[ f(x) = \frac{x - 1}{x + 1} \]

We can rearrange this to express \( x \) in terms of \( f(x) \):

\[ f(x)(x + 1) = x - 1 \]

Expanding this gives:

\[ xf(x) + f(x) = x - 1 \]

Rearranging terms:

\[ xf(x) - x = -1 - f(x) \]

Factoring out \( x \):

\[ x(f(x) - 1) = -1 - f(x) \]

Thus,

\[ x = \frac{-1 - f(x)}{f(x) - 1} \]

### Step 3: Substitute \( x \) into \( f(2x) \)

Now, we substitute \( x \) back into the expression for \( f(2x) \):

\[ f(2x) = f\left(2 \cdot \frac{-1 - f(x)}{f(x) - 1}\right) \]

Calculating \( 2x \):

\[ 2x = 2 \cdot \frac{-1 - f(x)}{f(x) - 1} = \frac{-2 - 2f(x)}{f(x) - 1} \]

Now substituting \( 2x \) into \( f \):

\[ f(2x) = \frac{\left(\frac{-2 - 2f(x)}{f(x) - 1}\right) - 1}{\left(\frac{-2 - 2f(x)}{f(x) - 1}\right) + 1} \]

### Step 4: Simplify the expression

The numerator becomes:

\[ \frac{-2 - 2f(x) - (f(x) - 1)}{f(x) - 1} = \frac{-2 - 2f(x) - f(x) + 1}{f(x) - 1} = \frac{-3f(x) - 1}{f(x) - 1} \]

The denominator becomes:

\[ \frac{-2 - 2f(x) + (f(x) - 1)}{f(x) - 1} = \frac{-2 - 2f(x) + f(x) - 1}{f(x) - 1} = \frac{-f(x) - 3}{f(x) - 1} \]

Thus, we have:

\[ f(2x) = \frac{\frac{-3f(x) - 1}{f(x) - 1}}{\frac{-f(x) - 3}{f(x) - 1}} = \frac{-3f(x) - 1}{-f(x) - 3} \]

### Step 5: Rearranging the expression

This simplifies to:

\[ f(2x) = \frac{3f(x) + 1}{f(x) + 3} \]

### Conclusion

Thus, we have proved that:

\[ f(2x) = \frac{3f(x) + 1}{f(x) + 3} \]