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Câu hỏi: Cho
\(\begin{array}{l}A = \left( {0; 4} \right), B = \left\{ {x \in R|{x^2} < 32} \right\},\\C = \left\{ {x \in R|{{\left({x - 2} \right)}^2} > 3} \right\}, D = \left[ {\sqrt 6 ; + \infty } \right)\end{array}\)
Tìm \(\left( {A \cap B} \right) \cup \left({C \cap D} \right)\)
\(\begin{array}{l}A = \left( {0; 4} \right), B = \left\{ {x \in R|{x^2} < 32} \right\},\\C = \left\{ {x \in R|{{\left({x - 2} \right)}^2} > 3} \right\}, D = \left[ {\sqrt 6 ; + \infty } \right)\end{array}\)
Tìm \(\left( {A \cap B} \right) \cup \left({C \cap D} \right)\)
Lời giải chi tiết
\(A\left( {0; 4} \right); B = \left({ - 4\sqrt 2; 4\sqrt 2 } \right).\) Vậy \(A \cap B = \left( {0; 4} \right)\)
\(\begin{array}{l}C = \left( { - \infty; 2 - \sqrt 3 } \right) \cup \left({2 + \sqrt 3 ; + \infty } \right);\\D = \left[ {\sqrt 6 ; + \infty } \right).\end{array}\)
Vậy \(C \cap D = \left( {2 + \sqrt 3 ; + \infty } \right)\)
Vậy \(\left( {A \cap B} \right) \cup \left({C \cap D} \right) = \left({0; + \infty } \right)\).
\(A\left( {0; 4} \right); B = \left({ - 4\sqrt 2; 4\sqrt 2 } \right).\) Vậy \(A \cap B = \left( {0; 4} \right)\)
\(\begin{array}{l}C = \left( { - \infty; 2 - \sqrt 3 } \right) \cup \left({2 + \sqrt 3 ; + \infty } \right);\\D = \left[ {\sqrt 6 ; + \infty } \right).\end{array}\)
Vậy \(C \cap D = \left( {2 + \sqrt 3 ; + \infty } \right)\)
Vậy \(\left( {A \cap B} \right) \cup \left({C \cap D} \right) = \left({0; + \infty } \right)\).