| √100 * 5x + 5 | ||
5x + 5x−2 + 5x−4 + ... = | ||
| 24 |
| 5 | √100 * 5x + 5 | ||
* 5x = | / * 24 | ||
| 4 | 24 |
| 5 | |
* 5x = √100 * 5x + 5 /2 | |
| 6 |
| 25 | |
52x = 100 * 5x + 5 5x = t > 0 licz dalej sama | |
| 36 |
| a1 | ||
S = | ||
| 1 − q |
| 1 | 1 | |||
a1 = 5x, q = | (myślałem że | ) W ramach rekompensacji masz porządne | ||
| 25 | 5 |
:
| √100 * 5x + 5 | ||
5x + 5x − 2 + 5x − 4 + ... = | } | |
| 24 |
| 5x | √100 * 5x + 5 | |||||||||||
= | } 5x = t | |||||||||||
| 24 |
| t | √100t + 5 | ||||||||
= | / * 24 | ||||||||
| 24 |
| 1 | 1 | |||
Po wyliczeni delt itd wyjdzie Ci − | − to odpada i | |||
| 25 | 5 |
| 1 | ||
t = | ||
| 5 |
| 1 | ||
5x = | ⇒ x = −1 | |
| 5 |
