If `(2)/(3)=(x-(1)/(y))+(x^(2)-(1)/(y^(2)))+ ... "To" oo`
and xy =2 then calculate the values of x and y with the condition that x `lt 1`.

Answer

To solve the problem, we need to find the values of \( x \) and \( y \) given the equation:

\[ \frac{2}{3} = \left( x - \frac{1}{y} \right) + \left( x^2 - \frac{1}{y^2} \right) + \left( x^3 - \frac{1}{y^3} \right) + \ldots \]

and the condition \( xy = 2 \) with \( x < 1 \).

### Step 1: Identify the series as a geometric series (GP)

The series can be separated into two parts:

1. The first part: \( x + x^2 + x^3 + \ldots \)
2. The second part: \( -\left( \frac{1}{y} + \frac{1}{y^2} + \frac{1}{y^3} + \ldots \right) \)

Both of these are geometric series.

### Step 2: Sum of the first geometric series

The sum of the first geometric series \( x + x^2 + x^3 + \ldots \) can be calculated using the formula for the sum of an infinite geometric series:

\[ S_1 = \frac{x}{1 - x} \quad \text{(where the first term is } x \text{ and the common ratio is } x) \]

### Step 3: Sum of the second geometric series

The sum of the second geometric series \( \frac{1}{y} + \frac{1}{y^2} + \frac{1}{y^3} + \ldots \) is:

\[ S_2 = \frac{\frac{1}{y}}{1 - \frac{1}{y}} = \frac{1/y}{(y-1)/y} = \frac{1}{y - 1} \quad \text{(where the first term is } \frac{1}{y} \text{ and the common ratio is } \frac{1}{y}) \]

### Step 4: Combine the sums

Now we can combine these sums into the original equation:

\[ \frac{2}{3} = \frac{x}{1 - x} - \frac{1}{y - 1} \]

### Step 5: Substitute \( y \) using \( xy = 2 \)

From the condition \( xy = 2 \), we can express \( y \) in terms of \( x \):

\[ y = \frac{2}{x} \]

Substituting this into the equation gives:

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

### Step 6: Simplify the equation

Substituting \( y \) into the second part of the equation:

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

### Step 7: Find a common denominator

The common denominator for the right-hand side is \( (1 - x)(2 - x) \):

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

### Step 8: Expand and simplify

Expanding the numerator:

\[ x(2 - x) - x(1 - x) = 2x - x^2 - x + x^2 = x \]

Thus, we have:

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

### Step 9: Cross-multiply

Cross-multiplying gives:

\[ 2(1 - x)(2 - x) = 3x \]

### Step 10: Expand and rearrange

Expanding the left side:

\[ 2(2 - 3x + x^2) = 3x \]
\[ 4 - 6x + 2x^2 = 3x \]
\[ 2x^2 - 9x + 4 = 0 \]

### Step 11: Solve the quadratic equation

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

\[ x = \frac{9 \pm \sqrt{(-9)^2 - 4 \cdot 2 \cdot 4}}{2 \cdot 2} \]
\[ x = \frac{9 \pm \sqrt{81 - 32}}{4} \]
\[ x = \frac{9 \pm \sqrt{49}}{4} \]
\[ x = \frac{9 \pm 7}{4} \]

This gives us two potential solutions for \( x \):

1. \( x = \frac{16}{4} = 4 \)
2. \( x = \frac{2}{4} = \frac{1}{2} \)

### Step 12: Determine the valid solution

Since we are given that \( x < 1 \), we take \( x = \frac{1}{2} \).

### Step 13: Calculate \( y \)

Using \( xy = 2 \):

\[ y = \frac{2}{x} = \frac{2}{\frac{1}{2}} = 4 \]

### Final Answer

Thus, the values of \( x \) and \( y \) are:

\[ x = \frac{1}{2}, \quad y = 4 \]