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Exerci Cio Su 01
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Exerci Cio Su 01
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Welton Oliveira
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∅
∈
∅
∅
A
⊂
∅
∅
∈{ } ∅
∅
⊂{ } ∅
∪ B = B ⇔ A ⊂ B ⇔ A ∩ B = A
⊂ U
A, B
C
= A C
C
= A C
∪ B) (A ∩ B ) (A
∩B ∪B
C C
P Q R U A B C Q R R A
C
∩ B ⊂ C
AC
C
⊂ B ∪ C
U
P P Q A B
AC A
C
∪ B ⊂ C C
C
⊂ B ∪ C
AC
C C
∪ B ⊂ C
\
{
∈
∈ }
B A = x ; x B e x / A .
\ ∅ B \ A = B B A =
⊂ A A ∩ B = ∅
B
\
\
B A = A B
\ \
A = B
\ \
A (B C ) = (A B ) C.
√ x + x = 2
√ x + 3 = x 2
⇔ x − 2x + 1 = 0 ⇔ x − 2 · 1 + 1 = 0 ⇔ x − 1 = 0 ⇔ x = ±1 x = 1 ⇔ x = ±1 x = 1
2 2
A B
C
A
∪ (B ∩ C ) = (A ∪ B) ∩ C. P (x)
x P (x) P (x)
x x A
x P (x)
U
x n x x
x2 = 1 n2 > n x > 1 x2 < 1
−
n
n>x n n>x
F M C P
x
4
P (A)
A A f : A
A
f
∈
}
A ; x / f (x)
→ P (A) { ∈
X = x
×
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