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Cartesian Product

A - B ∩ C = A...

Question

A-(B∩C) = A∩(B∩C)' = _________ = (A∩B')∪(A∩C') = (A-B)∪(A-C)

A∩(B'∪C')

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A(B'∪C')

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(A∩B')C'

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A∩(B'-C')

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Solution

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Similar questions

A \cap (B - C) = (A \cap B) - (A \cap C)

A \cap (B - C) = (A \cap B) - C

A \cup (B - C) = (A \cup B) \cap (A \cup C')

A \cup (B - C) = (A \cup B) - (A \cup C) .

A \cup (B \cap C) = (A \cup B) \cap (A \cup C)

A \cap (B \cup C) = (A \cap B) \cup (A \cap C)

A \cap (B - C) = (A \cap B) - (A \cap C)

A - (B \cup C) = A (A - B) \cap (A - C)

A - (B \cap C) = (A - B) \cup (A - C)

A \cap (B ∆ C) = (A \cap B) ∆ (A \cap C)

∣ ∣ ∣ ∣ ∣ \begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} - b c + c a + a b & b c - c a + a b & b c + c a - a b (a + b) (a + c) & (b + c) (b + a) & (c + a) (c + b) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 3. (b - c) (c - a) (a - b) (a + b + c) (a b + b c + c a)

∣ ∣ ∣ ∣ ∣ \begin{matrix} b c - a^{2} & c a - b^{2} & a b - c^{2} - b c + c a + c b & b c - c a + a b & b c + c a - a b (a + b) (a + c) & (b + c) (b + a) & (c + a) (c + b) \end{matrix} ∣ ∣ ∣ ∣ ∣ = 3 (b - c) (c - a) (a - b) (a + b + c) (b c + c a + a b)

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