| 3 | 6 | |||
a) | + | = | ||
| x | x−2 |
| 1 | 1 | |||
b) | + | = | ||
| x−1 | x+3 |
| x−2 | x−3 | |||
c) | + | = | ||
| x−1 | 1−x |
| 4 | 1−4x | |||
d) | + | = | ||
| x+3 | x2 + 3x |
| a | c | ad−cb | ||||
np. | − | = | i wykonujesz działania | |||
| b | d | bd |
| 4 | 1−4x | 4(x2+3x) + (1−4x)(x+3) | |||
+ | = | = | |||
| x+3 | x2+3x | (x+3)(x2+3x) |
| 4x2+12x + x+3−4x2−12x | ||
= | = | |
| (x+3)x(x+3 |
| x+3 | 1 | |||
= | = | |||
| (x+3)(x+3)x | x(x+3) |
tak to idzie? to d w sumie zrobilam tak jak Ty tylko na koncu wyszlo mi 3x a Tobie 1
dziekuje
za odp
dobrze robie?
| 3(x−2) | 3x−6+6x | |||
a) | + | |||
| x(x−2) | x(x−2) |
| x+3 | x−1 | x+3+x−1 | ||||
b) | + | = | = | |||
| (x−1) (x+3) | (x−1)(x+3) | (x−1)(x+3) |
| 2x+2 | ||
| (x−1)(x+3) |