`f(x)=(x-[x])/(1+x-[x])` and the number of different values which [f(x)] takes is `(lamda-1)` then `lamda` is ________ (where [.] denotes integer functions)

Answer

To solve the problem, we need to analyze the function \( f(x) = \frac{x - [x]}{1 + x - [x]} \), where \([x]\) denotes the greatest integer less than or equal to \(x\) (the floor function).

### Step 1: Understanding the Function
The term \(x - [x]\) represents the fractional part of \(x\), denoted as \(\{x\}\). Therefore, we can rewrite the function as:
\[ f(x) = \frac{\{x\}}{1 + \{x\}} \]

### Step 2: Analyzing the Range of \(\{x\}\)
The fractional part \(\{x\}\) takes values in the range:
\[ 0 \leq \{x\} < 1 \]

### Step 3: Finding the Range of \(f(x)\)
Next, we need to determine the range of \(f(x)\) as \(\{x\}\) varies from 0 to just below 1. We can analyze the behavior of \(f(x)\) at the endpoints of this interval.

- When \(\{x\} = 0\):
\[ f(0) = \frac{0}{1 + 0} = 0 \]

- As \(\{x\}\) approaches 1 (but does not reach it):
\[ f(\{x\} \to 1) = \frac{1}{1 + 1} = \frac{1}{2} \]

### Step 4: Continuous Function
Since \(f(x)\) is continuous in the interval \(0 \leq \{x\} < 1\), it will take all values between \(0\) and \(\frac{1}{2}\).

### Step 5: Finding the Values of \([f(x)]\)
Now we need to find the integer part \([f(x)]\). The function \(f(x)\) ranges from \(0\) to \(\frac{1}{2}\), which means:
\[ 0 \leq f(x) < \frac{1}{2} \]
Thus, the integer part \([f(x)]\) can only take the value:
\[ [f(x)] = 0 \]

### Step 6: Conclusion
Since \([f(x)]\) takes only one distinct value (which is 0), we can conclude that the number of different values that \([f(x)]\) takes is \(1\).

Given that the problem states that the number of different values which \([f(x)]\) takes is \((\lambda - 1)\), we set:
\[ \lambda - 1 = 1 \implies \lambda = 2 \]

### Final Answer
Thus, the value of \(\lambda\) is:
\[ \boxed{2} \]