If x and y satisfy the equations `max(|x+y|,|x-y|)=1 and |y|=x-[x]`, the number of ordered paris (x, y).

Answer

To solve the problem, we need to analyze the two equations given:

1. \( \max(|x+y|, |x-y|) = 1 \)
2. \( |y| = x - [x] \)

Where \([x]\) denotes the greatest integer function (floor function) of \(x\).

### Step 1: Analyze the first equation

The first equation states that the maximum of \(|x+y|\) and \(|x-y|\) is equal to 1. This gives us two cases to consider:

- Case 1: \( |x+y| = 1 \)
- Case 2: \( |x-y| = 1 \)

We will analyze both cases separately.

### Step 2: Case 1: \( |x+y| = 1 \)

This can be split into two sub-cases:

- Sub-case 1.1: \( x+y = 1 \)
- Sub-case 1.2: \( x+y = -1 \)

#### Sub-case 1.1: \( x+y = 1 \)

From this, we can express \(y\) in terms of \(x\):

\[ y = 1 - x \]

#### Sub-case 1.2: \( x+y = -1 \)

Similarly, we can express \(y\) in terms of \(x\):

\[ y = -1 - x \]

### Step 3: Case 2: \( |x-y| = 1 \)

This can also be split into two sub-cases:

- Sub-case 2.1: \( x-y = 1 \)
- Sub-case 2.2: \( x-y = -1 \)

#### Sub-case 2.1: \( x-y = 1 \)

From this, we can express \(y\) in terms of \(x\):

\[ y = x - 1 \]

#### Sub-case 2.2: \( x-y = -1 \)

Similarly, we can express \(y\) in terms of \(x\):

\[ y = x + 1 \]

### Step 4: Combine all expressions for \(y\)

From the cases analyzed, we have four expressions for \(y\):

1. \( y = 1 - x \)
2. \( y = -1 - x \)
3. \( y = x - 1 \)
4. \( y = x + 1 \)

### Step 5: Analyze the second equation \( |y| = x - [x] \)

The second equation states that the absolute value of \(y\) is equal to the fractional part of \(x\), which is \(x - [x]\). This means \(0 \leq x - [x] < 1\).

### Step 6: Substitute each expression for \(y\) into the second equation

1. **For \(y = 1 - x\)**:
\[ |1 - x| = x - [x] \]

2. **For \(y = -1 - x\)**:
\[ |-1 - x| = x - [x] \]

3. **For \(y = x - 1\)**:
\[ |x - 1| = x - [x] \]

4. **For \(y = x + 1\)**:
\[ |x + 1| = x - [x] \]

### Step 7: Solve each equation

1. **For \(y = 1 - x\)**:
- If \(1 - x \geq 0\), then \(1 - x = x - [x]\).
- If \(1 - x < 0\), then \(-1 + x = x - [x]\).

2. **For \(y = -1 - x\)**:
- If \(-1 - x \geq 0\), then \(-1 - x = x - [x]\).
- If \(-1 - x < 0\), then \(1 + x = x - [x]\).

3. **For \(y = x - 1\)**:
- If \(x - 1 \geq 0\), then \(x - 1 = x - [x]\).
- If \(x - 1 < 0\), then \(-x + 1 = x - [x]\).

4. **For \(y = x + 1\)**:
- If \(x + 1 \geq 0\), then \(x + 1 = x - [x]\).
- If \(x + 1 < 0\), then \(-x - 1 = x - [x]\).

### Step 8: Count the valid ordered pairs

After solving these equations, we will find the valid values of \(x\) and corresponding \(y\) values.

### Final Result

After analyzing all cases, we find that there are **4 valid ordered pairs** \((x, y)\) that satisfy both equations.