| 2 | |
≤ √ab
| |
| 1a+ 1b |
| 2 | 3 | ||
− | |||
| x2 − x | x2 + x |
| x3 + 5x2 − x − 5 | |
| x2 + 4x − 5 |
| 2 | |||||||
≤√ab | |||||||
|
| 2ab | |
≤√ab/()2 | |
| b+a |
| 4a2b2 | |
≤ab | |
| (a+b)2 |
| 4a2b2 | ab(a+b)2) | ||
− | ≤0 | ||
| (a+b)2 | (a+b)2 |
| 4a2b2−ab(a+b)2 | |
≤0 | |
| (a+b)2 |
| ab(4ab−a2−2ab−b2) | |
≤0 | |
| (a+b)2 |
| −ab(a−b)2 | |
≤0 | |
| (a+b)2 |
| −ab(a−b)2 | ||
−ab<0 więc iloraz | ≤0 | |
| (a+b)2 |
| 2 | 3 | 2(x+1)−3(x−1) | −x+4 | ||||
− | = | = | |||||
| x(x−1) | x(x+1) | x(x−1)(x+1) | x(x2−1) |
| x2(x+5)−(x+5) | (x+5)(x2−1) | (x−1)(x+1) | |||
= | = | =x+1
| |||
| (x−1)(x+5) | (x−1)(x+5) | x−1 |