(а+в)²+(а-в)² ______________ а²+в²

To simplify the expression \(\frac{(a+b)^2 + (a-b)^2}{a^2 + b^2}\), let's first simplify the numerator.

1. Expand \((a+b)^2\) and \((a-b)^2\):

   \[ (a + b)^2 = a^2 + 2ab + b^2 \] \[ (a - b)^2 = a^2 - 2ab + b^2 \]

2. Add the expanded forms:

   \[ (a+b)^2 + (a-b)^2 = (a^2 + 2ab + b^2) + (a^2 - 2ab + b^2) \] \[ = a^2 + 2ab + b^2 + a^2 - 2ab + b^2 \] \[ = 2a^2 + 2b^2 \]

3. Now, we can simplify the entire expression:

   \[ \frac{(a+b)^2 + (a-b)^2}{a^2 + b^2} = \frac{2a^2 + 2b^2}{a^2 + b^2} \]

4. Factor out the 2 in the numerator:

   \[ = \frac{2(a^2 + b^2)}{a^2 + b^2} \]

5. Since \(a^2 + b^2\) is not zero (assuming \(a\) and \(b\) are not both zero), we can cancel out \(a^2 + b^2\):

   \[ = 2 \]

Thus, the simplified expression is:

\[ \frac{(a+b)^2 + (a-b)^2}{a^2 + b^2} = 2 \]