Giải giúp e vs aaaaaaaaa

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Giải giúp e vs aaaaaaaaa

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8 tháng 2020-10-31T22:56:57+00:00 1 Answers 58 views 0

$\begin{array}{l} a)\left( {\dfrac{{\sqrt {14} – \sqrt 7 }}{{1 – \sqrt 2 }} + \dfrac{{\sqrt {15} – \sqrt 5 }}{{1 – \sqrt 3 }}} \right):\dfrac{1}{{\sqrt 7 – \sqrt 5 }}\\ = \left( {\dfrac{{ – \sqrt 7 \left( {1 – \sqrt 2 } \right)}}{{1 – \sqrt 2 }} + \dfrac{{ – \sqrt 5 \left( {1 – \sqrt 3 } \right)}}{{1 – \sqrt 3 }}} \right).\left( {\sqrt 7 – \sqrt 5 } \right)\\ = \left( { – \sqrt 7 – \sqrt 5 } \right).\left( {\sqrt 7 – \sqrt 5 } \right)\\ = – \left( {\sqrt 7 + \sqrt 5 } \right)\left( {\sqrt 7 – \sqrt 5 } \right)\\ = – \left( {7 – 5} \right)\\ = – 2\\ b)\left( {\dfrac{{2\sqrt 3 – \sqrt 6 }}{{\sqrt 8 – 2}} – \dfrac{{\sqrt {216} }}{3}} \right).\dfrac{1}{{\sqrt 6 }}\\ = \left( {\dfrac{{\sqrt 6 \left( {\sqrt 2 – 1} \right)}}{{2\sqrt 2 – 2}} – \dfrac{{\sqrt {36.6} }}{3}} \right).\dfrac{1}{{\sqrt 6 }}\\ = \left( {\dfrac{{\sqrt 6 \left( {\sqrt 2 – 1} \right)}}{{2\left( {\sqrt 2 – 1} \right)}} – \dfrac{{6\sqrt 6 }}{3}} \right).\dfrac{1}{{\sqrt 6 }}\\ = \left( {\dfrac{{\sqrt 6 }}{2} – 2\sqrt 6 } \right).\dfrac{1}{{\sqrt 6 }}\\ = \left( {\dfrac{1}{2} – 2} \right).\sqrt 6 .\dfrac{1}{{\sqrt 6 }}\\ = \dfrac{1}{2} – 2\\ = \dfrac{{ – 3}}{2} = – 1,5 \end{array}$