28.2. $$ 1-\frac{2 x y}{x^{2}+y^{2}} $$ 28.3. Find the product of 28.1 and 28.2 29. $$ \frac{4 x^{2}-4 x+1}{2 x^{2}+5 x-3} \div \frac{6 x^{2}-5 x+1}{3 x^{2}+8 x-3} $$

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**28.2. Simplify the Expression**

We start with
$$
1-\frac{2xy}{x^2+y^2}.
$$
Write 1 as a fraction with denominator $$x^2+y^2$$:
$$
1=\frac{x^2+y^2}{x^2+y^2}.
$$
Thus,
$$
1-\frac{2xy}{x^2+y^2}=\frac{x^2+y^2}{x^2+y^2}-\frac{2xy}{x^2+y^2}=\frac{x^2+y^2-2xy}{x^2+y^2}.
$$
Notice that
$$
x^2+y^2-2xy=(x-y)^2.
$$
So the simplified expression is
$$
\frac{(x-y)^2}{x^2+y^2}.
$$

**28.3. Product of 28.1 and 28.2**

Although the statement of 28.1 is not given explicitly, a common companion to the expression in 28.2 is
$$
1+\frac{2xy}{x^2+y^2}.
$$
We will assume that
$$
28.1: \quad 1+\frac{2xy}{x^2+y^2}.
$$
Then we may write it as
$$
1+\frac{2xy}{x^2+y^2}=\frac{x^2+y^2+2xy}{x^2+y^2}=\frac{(x+y)^2}{x^2+y^2}.
$$
Now, their product is
$$
\frac{(x+y)^2}{x^2+y^2}\cdot\frac{(x-y)^2}{x^2+y^2}=\frac{(x+y)^2(x-y)^2}{(x^2+y^2)^2}.
$$
Since
$$
(x+y)^2(x-y)^2=\bigl[(x+y)(x-y)\bigr]^2=(x^2-y^2)^2,
$$
the final result is
$$
\frac{(x^2-y^2)^2}{(x^2+y^2)^2}.
$$

**29. Divide and Simplify**

We need to simplify
$$
\frac{4x^2-4x+1}{2x^2+5x-3} \div \frac{6x^2-5x+1}{3x^2+8x-3}.
$$

_Step 1. Factor all quadratics._

1. For the numerator of the first fraction:
   $$
   4x^2-4x+1=(2x-1)^2.
   $$
2. For the denominator of the first fraction:
   Find numbers that multiply to $$2(-3)=-6$$ and add to $$5$$. We have:
   $$
   2x^2+6x-x-3= (2x^2+6x)- (x+3)=2x(x+3)-1(x+3)=(2x-1)(x+3).
   $$
   Thus,
   $$
   2x^2+5x-3=(2x-1)(x+3).
   $$
3. For the numerator of the second fraction:
   We factor
   $$
   6x^2-5x+1.
   $$
   Notice that
   $$
   (2x-1)(3x-1)=6x^2-2x-3x+1=6x^2-5x+1.
   $$
4. For the denominator of the second fraction:
   Factor
   $$
   3x^2+8x-3.
   $$
   Write it as:
   $$
   3x^2+9x-x-3= 3x(x+3)-1(x+3)=(3x-1)(x+3).
   $$
   
_Step 2. Rewrite the whole expression using the factors:_

The expression becomes
$$
\frac{(2x-1)^2}{(2x-1)(x+3)} \div \frac{(2x-1)(3x-1)}{(3x-1)(x+3)}.
$$

_Step 3. Change division to multiplication by the reciprocal:_
$$
\frac{(2x-1)^2}{(2x-1)(x+3)} \times \frac{(3x-1)(x+3)}{(2x-1)(3x-1)}.
$$

_Step 4. Cancel common factors:_

- $$x+3$$ cancels.
- $$3x-1$$ cancels.
- In the factors of $$2x-1$$: There is one factor in the first denominator and one in the second denominator. The numerator has $$(2x-1)^2$$. Cancel one $$2x-1$$ from the numerator with the one in the first denominator, and then cancel the remaining $$2x-1$$ from the numerator with the one in the second denominator.

After canceling, we are left with
$$
1.
$$

Thus,
$$
\frac{4x^2-4x+1}{2x^2+5x-3} \div \frac{6x^2-5x+1}{3x^2+8x-3} = 1.
$$
**28.2. Simplified Expression:** $$ \frac{(x - y)^2}{x^2 + y^2} $$ **28.3. Product of 28.1 and 28.2:** $$ \frac{(x^2 - y^2)^2}{(x^2 + y^2)^2} $$ **29. Simplified Division:** $$ 1 $$

28.2. \( 1-\frac{2 x y}{x^{2}+y^{2}} \) 28.3. Find the product of 28.1 and 28.2 29. \( \frac{4 x^{2}-4 x+1}{2 x^{2}+5 x-3} \div \frac{6 x^{2}-5 x+1}{3 x^{2}+8 x-3} \)

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