1 | ||
2bd = b ⇔ 2d=1 ⇔ d= | ||
2 |
1 | ||
czyli element ep = | *i | |
2 |
1 | ||
2bd = d ⇔ 2b=1 ⇔ b= | ||
2 |
1 | ||
czyli el = | *i | |
2 |
1 | ||
el=ep czyli mamy element neutralny e= | *i | |
2 |
1 | ||
(3−3i)o(c+di) = | *i | |
2 |
1 | ||
(3+c)+2*(−3)*d*i = | *i | |
2 |
1 | ||
−6d= | ||
2 |
1 | ||
d = − | ||
12 |
1 | ||
element odwrotny do 3−3i to −3− | i | |
12 |
5^2 | 52 |
2^{10} | 210 |
a_2 | a2 |
a_{25} | a25 |
p{2} | √2 |
p{81} | √81 |
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