| 3a+3−2*3a+z | ||
zd27. Wykaż,że wyrażenie | jest liczbą wymierną. | |
| 3a+1 +3a−1 |
| 3a+2+1−2*3a+2 | 3a+2*31 − 2*3a+2 | |||
= | = | = | ||
| 3a−1+2+3a−1 | 3a−1*32+3a−1 |
| 3a+2*3−2*3a+2 | 3a+2*(3−2) | |||
= | = | = | ||
| 3a−1*9+3a−1 | 3a−1*(9+1) |
| 1*3a+2 | 3a+2 | 1 | 1 | |||||
= | = | * | = 3a+2−(a−1)* | = | ||||
| 10*3a−1 | 3a−1 | 10 | 10 |
| 1 | 1 | 27 | ||||
= 3a+2−a+1* | = 33* | = | ∊ W | |||
| 10 | 10 | 10 |
| 3a+3−2*3a+z | 33−2*3z | ||
= | = | ||
| 3a+1+3a−1 | 3+3−1 |
| 81−6*3z | ||
= | ||
| 10 |
| 1 | 1 | 81−6*3z | −√2 | |||||
weźmy z=log3( | √2+ | *81) wtedy mamy | = | ∊ℛ\ℚ | ||||
| 6 | 6 | 10 | 10 |