If `x= (sqrt87- sqrt71)/(sqrt55 + sqrt39) " & "y= (sqrt87 + sqrt71)/(sqrt55- sqrt39)` then find `(1)/(x+1) + (1)/(y+ 1)=` ?

Answer

To solve the problem, we need to find the value of \( \frac{1}{x+1} + \frac{1}{y+1} \) given:

\[ x = \frac{\sqrt{87} - \sqrt{71}}{\sqrt{55} + \sqrt{39}} \]
\[ y = \frac{\sqrt{87} + \sqrt{71}}{\sqrt{55} - \sqrt{39}} \]

### Step 1: Calculate \( x \cdot y \)

We start by calculating the product \( x \cdot y \):

\[ x \cdot y = \left( \frac{\sqrt{87} - \sqrt{71}}{\sqrt{55} + \sqrt{39}} \right) \cdot \left( \frac{\sqrt{87} + \sqrt{71}}{\sqrt{55} - \sqrt{39}} \right) \]

Using the difference of squares formula \( (a-b)(a+b) = a^2 - b^2 \):

\[ = \frac{(\sqrt{87})^2 - (\sqrt{71})^2}{(\sqrt{55} + \sqrt{39})(\sqrt{55} - \sqrt{39})} \]

Calculating the squares:

\[ = \frac{87 - 71}{55 - 39} \]

This simplifies to:

\[ = \frac{16}{16} = 1 \]

### Step 2: Calculate \( x + y \)

Next, we need to find \( x + y \):

\[ x + y = \frac{\sqrt{87} - \sqrt{71}}{\sqrt{55} + \sqrt{39}} + \frac{\sqrt{87} + \sqrt{71}}{\sqrt{55} - \sqrt{39}} \]

Finding a common denominator:

\[ = \frac{(\sqrt{87} - \sqrt{71})(\sqrt{55} - \sqrt{39}) + (\sqrt{87} + \sqrt{71})(\sqrt{55} + \sqrt{39})}{(\sqrt{55} + \sqrt{39})(\sqrt{55} - \sqrt{39})} \]

Calculating the numerator:

\[ = \frac{(\sqrt{87} \sqrt{55} - \sqrt{87} \sqrt{39} - \sqrt{71} \sqrt{55} + \sqrt{71} \sqrt{39}) + (\sqrt{87} \sqrt{55} + \sqrt{87} \sqrt{39} + \sqrt{71} \sqrt{55} + \sqrt{71} \sqrt{39})}{16} \]

Combining like terms:

\[ = \frac{2\sqrt{87} \sqrt{55} + 2\sqrt{71} \sqrt{39}}{16} \]

Thus, we can simplify:

\[ = \frac{\sqrt{87} \sqrt{55} + \sqrt{71} \sqrt{39}}{8} \]

### Step 3: Calculate \( \frac{1}{x+1} + \frac{1}{y+1} \)

Now, we can use the formula:

\[ \frac{1}{x+1} + \frac{1}{y+1} = \frac{y+1 + x+1}{(x+1)(y+1)} = \frac{x+y+2}{xy + x + y + 1} \]

Substituting the values we found:

- \( xy = 1 \)
- \( x + y = 2 \)

Thus:

\[ = \frac{2 + 2}{1 + 2 + 1} = \frac{4}{4} = 1 \]

### Final Answer

The final answer is:

\[ \frac{1}{x+1} + \frac{1}{y+1} = 1 \]