| 1 | a3+a | |||
log[x−a(1−a)−0,5]−0,5log(1+ | )−log√ | −a2( w ostatnim wyrazie calosc jest | ||
| a | a+1 |
| a2+a | a3+a−a2(a+1) | a−a2 | ||||
Rozpiszse tylko sobie na boku | −a2= | = | ||||
| a+1 | a+1 | a+1 |
| a | a+1 | a−a2 | ||||
log[x− | ]−log√ | −log√ | ||||
| √1−a | a | a+1 |
| a | a(1−a) | a+1 | ||||
log[x− | ]= log√ | +log√ | ||||
| √1−a | a+1 | a |
| a | a(1−a | a+1 | ||||
log[x− | =log√( | )*( | ) | |||
| √1−a | a+1 | a |
| a | ||
x− | =√1−a | |
| √1−a |
| a | ||
x=√1−a+ | ||
| √1−a |
| √1−a)2+a | ||
x=( | ||
| √1−a |
| 1−a+a | 1 | √1−a | ||||
x= | = | = | (chociaż moglem zostwic postac porzednia . | |||
| √1−a | √1−a | 1−a |