If `f(x)=(2010x+165)/(165x-2010), x gt 0 " and " x ne (2010)/(165)`, the least value of `f(f(x))+f(f(4/x))` is ………. .

Answer

To solve the problem, we need to find the least value of the expression \( f(f(x)) + f(f(4/x)) \) where \( f(x) = \frac{2010x + 165}{165x - 2010} \).

### Step 1: Find \( f(f(x)) \)

We start by substituting \( f(x) \) into itself.

\[ f(f(x)) = f\left(\frac{2010x + 165}{165x - 2010}\right) \]

Substituting \( f(x) \) into the function:

\[ f(f(x)) = \frac{2010\left(\frac{2010x + 165}{165x - 2010}\right) + 165}{165\left(\frac{2010x + 165}{165x - 2010}\right) - 2010} \]

### Step 2: Simplify \( f(f(x)) \)

Now we simplify the numerator and denominator separately.

**Numerator:**
\[ 2010\left(\frac{2010x + 165}{165x - 2010}\right) + 165 = \frac{2010(2010x + 165) + 165(165x - 2010)}{165x - 2010} \]

Calculating the numerator:
\[ 2010^2 x + 2010 \cdot 165 + 165 \cdot 165 x - 165 \cdot 2010 \]
\[ = (2010^2 + 165^2)x + (2010 \cdot 165 - 165 \cdot 2010) = (2010^2 + 165^2)x \]

**Denominator:**
\[ 165\left(\frac{2010x + 165}{165x - 2010}\right) - 2010 = \frac{165(2010x + 165) - 2010(165x - 2010)}{165x - 2010} \]

Calculating the denominator:
\[ 165 \cdot 2010 x + 165^2 - 2010 \cdot 165 x + 2010^2 = (165 \cdot 2010 - 2010 \cdot 165)x + 165^2 + 2010^2 = 165^2 + 2010^2 \]

Thus, we have:
\[ f(f(x)) = \frac{(2010^2 + 165^2)x}{165^2 + 2010^2} \]

### Step 3: Find \( f(f(4/x)) \)

Now we need to find \( f(f(4/x)) \):
\[ f(f(4/x)) = f\left(\frac{2010 \cdot \frac{4}{x} + 165}{165 \cdot \frac{4}{x} - 2010}\right) \]

Following the same process as above, we can find:
\[ f(f(4/x)) = \frac{(2010^2 + 165^2) \cdot \frac{4}{x}}{165^2 + 2010^2} \]

### Step 4: Combine \( f(f(x)) \) and \( f(f(4/x)) \)

Now we add \( f(f(x)) + f(f(4/x)) \):
\[ f(f(x)) + f(f(4/x)) = \frac{(2010^2 + 165^2)x}{165^2 + 2010^2} + \frac{(2010^2 + 165^2) \cdot \frac{4}{x}}{165^2 + 2010^2} \]

Factoring out the common term:
\[ = \frac{2010^2 + 165^2}{165^2 + 2010^2} \left( x + \frac{4}{x} \right) \]

### Step 5: Find the minimum value of \( x + \frac{4}{x} \)

To find the minimum value of \( x + \frac{4}{x} \), we can use the AM-GM inequality:
\[ x + \frac{4}{x} \geq 2\sqrt{x \cdot \frac{4}{x}} = 4 \]

### Conclusion

Thus, the least value of \( f(f(x)) + f(f(4/x)) \) is:
\[ \frac{2010^2 + 165^2}{165^2 + 2010^2} \cdot 4 \]

Since the fraction simplifies to 1, the least value is:
\[ \boxed{4} \]