| 1 | 1 | sina | cosa | |||||
( | − | )*(1+tga+ctga)= | − | |||||
| cosa | sina | cos2a | sin2a |
| sina | cosa | sina | cosa | |||||
L=( | − | )*(1+ | + | )= | ||||
| sinacosa | sinacosa | cosa | sina |
| sina−cosa | sin2+cos2 | |||
( | )*(1+ | )= | ||
| sinacosa | sinacosa |
| 1+sinacosa | ||
( to samo)*( | ).. dobrze liczę? | |
| sinacosa |
| 1 | 1 | a +1 | ||||
Rozwiązuję ten przyklad | + | = | i po sprowadzeniu do wspólnego | |||
| x − a | x − 1 | a |
| 1 | 1 | 2(a +3) | ||||
a) | − | = | ||||
| 2a + ax | 2x − x2 | x3 − 4x |
| 1 | 1 | a +1 | ||||
b) | + | = | ||||
| x − a | x − 1 | a |
| x2 +1 | 1 | x | ||||
c) | + | = | ||||
| a2 * x − 2a | ax − 2 | a |
| x3+3x2−2x−6 | |
| x2−2 |
| x2(x+3)−2(x+3) | |
| x2−2 |