| 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+4 | x−4 | x+8 | |||
− | = | ||||
| x2−4x | 2x2−8x | 2x2−32 |
| P1 | b2 | ||
= | P−pole trójkąta ABC | ||
| P | (a+b)2 |
| P2 | a2 | ||
= | |||
| P | (a+b)2 |
| P2 | a2 | ||
= | |||
| P1 | b2 |
| bh1 | 2P1 | |||
P1= | ⇒ h1= | |||
| 2 | b |
| ah2 | ||
P2= | ||
| 2 |
| a | P2 | |||
Pole równoległoboku = ah1 = 2P1 | =2P1√ | |||
| b | P1 |
| x3−3x−2 | ||
m= | , z czego wynikałoby, że jedyną taką liczbą jest 2. | |
| x(x−2) |