| x2 | 64x−64 | |||
( | )2 = | |||
| x−1 | x2 |
| x2 | ||
t = | ||
| x−1 |
| 64(x−1) | ||
t = | ||
| x2 |
| x4 | 64*( x − 1) | ||
= | mnożymy na krzyż | ||
| ( x −1)2 | x2 |
| x−1 | 64 | |||
P = 64* | = | |||
| x2 | t |
| 64 | ||
masz równanie: t2 = | ⇔ t3 = 64 | |
| t |
| 1 | ||
t=64 | ||
| t |
| 1 | ||
przepraszam babol : t2=64 | ||
| t |
| x2 | ||
dzieki kix, tak wlasnie myslalam, to znaczy ze jesli t = | to | |
| x−1 |
| (x−1) | 1 | ||
= |  | ||
| x2 | t |
| x2 | |
= 4 ⇔ x = 2 | |
| x−1 |
| x−1 | 1 | |||
tak to nie było problemem tylko wlasnie niemoglam wpasc jak z tego | zrobic | |||
| x2 | t |
