Epyxoid
{ Tanár }
megoldása
3 éve
`U = {0; 1; 2; 3; 4; 5; 6; 7; 8; 9}", " A = {1; 3; 4; 5; 7}", " B = {2; 3; 4; 5; 6}", " C = {0; 2; 4; 6; 8; 9}`
a) `bar A nn bar B = {0; 2; 6; 8; 9} nn {0; 1; 7; 8; 9} = {0; 8; 9}`
b) `bar A nn B = {0; 2; 6; 8; 9} nn {2; 3; 4; 5; 6} = {2; 6}`
c) `bar(A nn B) = bar({3; 4; 5}) = {0, 1; 2; 6; 7; 8; 9}`
d) `bar(A uu B) = bar({1; 2; 3; 4; 5; 6; 7}) = {0; 8; 9}`
e) `bar A uu bar B = {0; 2; 6; 8; 9} uu {0; 1; 7; 8; 9} = {0; 1; 2; 6; 7; 8; 9}`
f) `A uu bar B = {1; 3; 4; 5; 7} uu {0; 1; 7; 8; 9} = {0, 1; 3; 4; 5, 7; 8; 9}`
g) `A nn bar B = {1; 3; 4; 5; 7} nn {0; 1; 7; 8; 9} = {1; 7}`
h) `A uu C = {1; 3; 4; 5; 7} uu {0; 2; 4; 6; 8; 9} = U`
i) `B nn C = {2; 3; 4; 5; 6} nn {0; 2; 4; 6; 8; 9} = {2; 4; 6}`
j) `bar B uu C = {0; 1; 7; 8; 9} uu {0; 2; 4; 6; 8; 9} = {0; 1; 2; 4; 6; 7; 8; 9}`
k) `(A nn B) uu C = {3; 4; 5} uu {0; 2; 4; 6; 8; 9} = {0; 2; 3; 4; 5; 6; 8; 9}`
l) `(A uu B) nn C = {1; 2; 3; 4; 5; 6; 7} nn {0; 2; 4; 6; 8; 9} = {2; 4; 6}`
m) `(bar A nn B) uu C = {2; 6} uu {0; 2; 4; 6; 8; 9} = C`
n) `(bar A uu B) nn bar C = ({0; 2; 6; 8; 9} uu {2; 3; 4; 5; 6}) nn {1; 3; 5; 7} ``=`` {0; 2; 3; 4; 5; 6; 8; 9} nn {1; 3; 5; 7} = {3; 5}`
o) `bar(A uu B) nn bar C = {0; 8; 9} nn {1; 3; 5; 7} = O/`
p) `bar A nn bar B nn bar C = {0; 8; 9} nn {1; 3; 5; 7} = O/`
Módosítva: 3 éve
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