| 1 | 3 | |||
1/ | + | ≥ (x−1)(x−2) | ||
| x−1 | x−2 |
| 1 | 3 | ||||||||
+ | ≥
| ||||||||
| x−2 |
| 1 | ||||||||||||||
czy | ≥ | |||||||||||||
|
| 1 | |||||||
≥(x−1)(x−2)
| |||||||
|
| (x−1)(x−2) | |
≥(x−1)(x−2)
| |
| 4x−5 |
| (x−1)(x−2) | |
−(x−1)(x−2)≥0
| |
| 4x−5 |
| 1−4x+5 | ||
(x−1)(x−2) | ≥0
| |
| 4x−5 |
| −4x+6 | ||
(x−1)(x−2) | ≥0 ...itd− ![]() | |
| 4x−5 |
| −4(x−1,5)(x−1)(x−2) | |
≥0 ... zamieniasz na iloczyn
| |
| 4(x−5/4) |
| 5 | ||
Dodam od siebie(jakby ktoś inny miał problem), że trzeba odrzucić x=1, x=2, x= | w | |
| 4 |
| 5 | 3 | |||
x∊(1, | ) U < | , 2) | ||
| 4 | 2 |