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Soal dan Pembahasan Ayo Kita Berlatih 2.9 Matematika kelas 7 Bab Himpunan K13


Soal dan Pembahasan Ayo Kita Berlatih 2.9 Matematika kelas 7 Bab Himpunan K13


Ayo Kita Berlatih 2.9 !
1). Diketahui:
S = {bilangan asli kurang dari 15}
P = {1, 2, 3, 4, 5, 6}
Q = {5, 6,7, 8, 9}
Tentukan
a. Pc
b. Qc
c. (P ∩ Q)c
d. (P ∪ Q)

Jawab :
*Pc = P'
Diketahui S = {bilangan asli kurang dari 15}, P = {1, 2, 3, 4, 5, 6}, dan Q ={5, 6, 7, 8, 9}. Tentukan anggota dari P', Q', (P ∩ Q)', dan (P ∪ Q)'!
S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14}
P = {1, 2, 3, 4, 5, 6}
Q ={5, 6, 7, 8, 9} 

a. P' = {7, 8, 9, 10, 11, 12, 13, 14}
b. Q' = {1, 2, 3, 4, 10, 11, 12, 13, 14}
c. P ∩ Q = {5, 6}
    P ∪ Q = {1, 2, 3, 4, 5, 6, 7, 8, 9}
    (P ∩ Q)' = {1, 2, 3, 4, 7, 8, 9, 10, 11, 12, 13, 14}
d. (P ∪ Q)' = {10, 11, 12, 13, 14}

2). Diketahui
S = {bilangan cacah kurang dari 11}
A = {x | x ∈ P, x < 10, P bilangan prima}
B = {5, 7, 9}
Tentukan
a. Ac
b. Bc
c. (A ∩ B)c
d. (A ∪ B)c
e. A ∩ (A ∪ B)c
f. Bc ∩ (A ∪ B)
g. (A ∪ B)c ∩ (A ∪ B)c
h. (Ac ∩ B)c ∪ (A ∪ B c)c

Jawab :
S = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
A = {2, 3, 5, 7}
B = {5, 7, 9}
a. A' = {0, 1, 4, 6, 8, 9, 10}. 

b. B' = {0, 1, 2, 3, 4, 6, 8, 10}.  
A ∩ B = {2, 3, 5, 7} ∩ {5, 7, 9} = {5, 7}. 

c. (A ∩ B)' = {0, 1, 2, 3, 4, 6, 8, 9, 10}.
A ∪ B = {2, 3, 5, 7} ∪ {5, 7, 9} = {2, 3, 5, 7, 9}.
 
d. (A ∪ B)' = {0, 1, 4, 6, 8, 10}. 

e. A ∩ (A U B)' = {2, 3, 5, 7} ∩ {0, 1, 4, 6, 8, 10} = ∅. 

f. B' ∩ (A ∪ B) = {0, 1, 2, 3, 4, 6, 8, 10} ∩ {2, 3, 5, 7, 9} = {2, 3}. 

g. (A ∪ B)' ∩ (A ∪ B)' = {0, 1, 4, 6, 8, 10} ∩ {0, 1, 4, 6, 8, 10} = {0, 1, 4, 6, 8, 10}.
A' ∩ B = {0, 1, 4, 6, 8, 9, 10} ∩ {5, 7, 9} = {9}.
(A' ∩ B)' = {0, 1, 2, 3, 4, 5, 6, 7, 8, 10}. 

h. (A' ∩ B)' ∪ (A ∪ B')' = {0, 1, 2, 3, 4, 5, 6, 7, 8, 10} ∪ {0, 1, 4, 6, 8, 10} = {0, 1, 4, 6, 8, 10}.


3). Perhatikan diagram Venn berikut ini 

Soal dan Pembahasan Ayo Kita Berlatih 2.9 Matematika kelas 7 Bab Himpunan K13

Berdasarkan diagram Venn tersebut tentukan banyak anggota dari
a. Ac
b. Bc
c. Cc
d. (A ∩ B)c
e. (A ∪ C)c
f. (A ∩ C)c
g. Ac ∩ (B ∪ C)c
h. (A ∩ B)c ∩ (A ∩ C)c

Jawab :
a. banyak anggota complemen A = 25 + 3 + 18 = 46 
b . banyak anggota complemen B = 20 + 5 + 18 = 43 
c. banyak anggota complemen C = 20 + 7 + 25 = 52 
d. banyak anggota dari complemen irisan A dan B = 20 + 5 + 18 + 3 + 25 = 71
e. banyak anggota dari complemen gabungan A dan C = 25
f. banyak anggota dari complemen irisan A dan C = 20 + 7 + 18 + 3 + 25 = 73
g. banyak anggota dari irisan complemen A dengan complemen gabungan B dan C = 0
h. banyak anggota dari irisan complemen irisan A dan B dengan complemen irisan A dan C 
= 20 + 18 + 3 + 25 = 66

4). Diketahui A = {a, b, c, d, e, f} dan B = {e, f, g, h, j}. Tentukan
a. A – B
b. B – A
c. (A – B) ∩ A
d. (A – B) ∪ (B – A)
5. Misalkan A = {1, 2, 3, ..., 10} dan B = {3, 5, 7, 9,11, 13}, dan
C = {7, 8, 9, 10, 11, 12, 13}. Tentukan anggota himpunan dari
a. A – B
b. B – A
c. B – C
d. C – A
e. (A – B) ∩ (A – C)
f. (A – C) ∪ (B – C)
g. (A ∪ B) – (B ∪ C)
h. (A – B)c ∪ (B – C)

Jawab :
a. A - B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {3, 5, 7, 9, 11, 13} = {1, 2, 4, 6, 8, 10}. 

b. B - A = {3, 5, 7, 9, 11, 13} - {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} = {11, 13}. 

c. B - C = {3, 5, 7, 9, 11, 13} - {7, 8, 9, 10, 11, 12, 13} = {3, 5}
   C - B = {7, 8, 9, 10, 11, 12, 13} - {3, 5, 7, 9, 11, 13} = {8, 10, 12} 

d. C - A = {7, 8, 9, 10, 11, 12, 13} - {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} = {11, 12, 13}
    A - C = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} - {7, 8, 9, 10, 11, 12, 13} = {1, 2, 3, 4, 5, 6} 

e. (A - B) ∩ (A - C) = {1, 2, 4, 6, 8, 10} ∩ {1, 2, 3, 4, 5, 6} = {1, 2, 4, 6}. 

f. (A - C) ∪ (B - C) = {1, 2, 3, 4, 5, 6} ∪ {3, 5} = {1, 2, 3, 4, 5, 6}
    A U B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} ∪ {3, 5, 7, 9, 11, 13} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13}
    B ∪ C = {3, 5, 7, 9, 11, 13} ∪ {7, 8, 9, 10, 11, 12, 13} = {3, 5, 7, 8, 9, 10, 11, 12, 13} 

g. (A∪B)-(B∪C) = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 13}-{3, 5, 7, 8, 9, 10, 11, 12, 13} = {1, 2, 4, 6, 12}
Untuk menentukan komplemen, seharusnya diketahui himpunan semesta S. Kemungkinan S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13} sehingga 
(A - B)' = {1, 2, 4, 6, 8, 10}' = {3, 5, 7, 9, 11, 13} 
(B - C)' = {3, 5}' = {1, 2, 4, 6, 7, 8, 9, 10, 11, 12, 13} 

h. (A - B)' ∪ (B - C)' = {3, 5, 7, 9, 11, 13} ∪ {1, 2, 4, 6, 7, 8, 9, 10, 11, 12, 13} = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13} = S.

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