| x2−16 | x2 +x | x3+4x2 | |||
x | : | ||||
| 1−x2 | 3x−12 | x−1 |
| (x−4)(x+4) | x(1+1) | x2(x+4) | ||||
D={−1,1,4}, a rozwiązanie to : | x | : | ||||
| (x+1)(x−1) | 3(x−4) | x−1 |
| (x+4) | x | (x−1) | x | |||||
= | x | x | = | ? | ||||
| (x−1) | 3 | x2(x+4) | 3x2 |
| x | 3 | |||
. | = | |||
| 3x2 | x |
, tylko 0 skąd?
| x3+4x2 | |
≠ 0 | |
| x−1 |
| 5^2 | 52 |
| 2^{10} | 210 |
| a_2 | a2 |
| a_{25} | a25 |
| p{2} | √2 |
| p{81} | √81 |
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