| x+1 | 2 | |||
a) | − | = mi wychodzi: −5x−x2(2−x)(3x+1) a powinno być: | ||
| 3x+1 | 2−x |
| x2+5x | ||
| (3x+1)(x−2) |
| x | x+1 | 2x+3 | −2 | 2 | ||||||
b) | − | − | = mi wychodzi | a powinno być: | ||||||
| x−4 | 4−x | x−4 | x−4 | 4−x |
| −5x−x2 | (−1)*(5x+x2) | x2+5x | |||
= | = | ||||
| (2−x)*(3x+1) | (−1)*(x−2)*(3x+1) | (x−2)*(3x+1) |
| −2 | −2 | 2 | |||
= | = | ||||
| x−4 | (−1)*(4−x) | 4−x |
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