| 2n2 | ||
( | )n−n2 | |
| 2n2−1 |
i:
| 2n2−1+1 | 1 | 1 | ||||
( | )n−n2 = (1+ | )n−n2 = [(1+ | )n−n2] −> i | |||
| 2n2−1 | 2n2−1 | 2n2−1 |
| n−n2 | ||
jeszcze do potegi |  ? | |
| 2n2−1 |
| n−n2 | ||
i ze potem to jest e do potegi | ||
| 2n2−1 |
prosze o pomoc 
| n−n2 | 1n − 1 | −1 | |||
= | → | ||||
| 2n2−1 | 2 − 1n | 2 |
| 1 | 1 | |||
e−1/2 = | = | |||
| e1/2 | √e |
| 1 | ||
a zeby wyszlo e to musi byc zawsze (1+ | )x | |
| x |