| (4x3−6x)(x2−1)2−(x4−3x2)(x2−1)*4x | |
| (x2−1)4 |
| 1 | 1 | |||
limx−>0+=( | − | ) | ||
| 1−cosx | x2 |
| 1 | ||
wiedzac ze sinα−cosα=− | oblicz sin2α | |
| 6 |
| 1 | ||
sinα−cosα=− | /2 | |
| 6 |
| 1 | ||
sin2α−2sinαcosα+cos2α= | ||
| 36 |
| 1 | ||
−sin2α=−1+ | ||
| 36 |
| 35 | ||
sin2α= | ||
| 36 |
| pi | ||
Dana jest liczba sin(cos | ). Zatem: | |
| 3 |
| √3 | pi | |||
A. | < sin(cos | ) < 1 | ||
| 2 | 3 |
| √2 | pi | √3 | ||||
B. | < sin(cos | ) < | ||||
| 2 | 3 | 2 |
| 1 | pi | √2 | ||||
C. | < sin(cos | ) < | ||||
| 2 | 3 | 2 |
| pi | 1 | |||
D. 0 < sin(cos | ) < | |||
| 3 | 2 |
| 5n+7 | ||
Dany jest zbiór A={ | : n ∊ N}. Korzystając z definicji uzasadnić, że supA = 5. | |
| n+2 |