Answer
**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:**
\[
1
\]
Solution
**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,
\[
\frac{4x^2-4x+1}{2x^2+5x-3} \div \frac{6x^2-5x+1}{3x^2+8x-3} = 1.
\]
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