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Rút gọn :
a) \(2\sqrt {28} + 3\sqrt {63} – 2\sqrt {112} – \sqrt {175} \);
b) \(\sqrt 2 .\sqrt {7 + 3\sqrt 5 } – \dfrac{4}{{\sqrt 5 – 1}}\);
c) \(\dfrac{1}{{2\sqrt 3 + 3}} – \dfrac{1}{{2\sqrt 3 – 3}}\);
d) \(\sqrt {27b} + 12\sqrt {\dfrac{1}{3}b} – \sqrt {48b} \left( {b > 0} \right)\).
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\(\begin{array}{l}a)\;2\sqrt {28} + 3\sqrt {63} – 2\sqrt {112} – \sqrt {175} \\ = 2\sqrt {{2^2}.7} + 3\sqrt {{3^2}.7} – 2\sqrt {{4^2}.7} – \sqrt {{5^2}.7} \\ = 2.2\sqrt 7 + 3.3\sqrt 7 – 2.4\sqrt 7 – 5\sqrt 7 \\ = 4\sqrt 7 + 9\sqrt 7 – 8\sqrt 7 – 5\sqrt 7 = 0.\end{array}\) \(\begin{array}{l}c)\;\;\dfrac{1}{{2\sqrt 3 + 3}} – \dfrac{1}{{2\sqrt 3 – 3}}\\ = \dfrac{{2\sqrt 3 – 3}}{{\left( {2\sqrt 3 + 3} \right)\left( {2\sqrt 3 – 3} \right)}} – \dfrac{{2\sqrt 3 + 3}}{{\left( {2\sqrt 3 + 3} \right)\left( {2\sqrt 3 – 3} \right)}}\\ = \dfrac{{2\sqrt 3 – 3 – 2\sqrt 3 – 3}}{{{{\left( {2\sqrt 3 } \right)}^2} – 9}}\\ = \dfrac{{ – 6}}{{12 – 9}} = \dfrac{{ – 6}}{3} = – 2.\end{array}\) |
\(\begin{array}{l}b)\;\sqrt 2 .\sqrt {7 + 3\sqrt 5 } – \dfrac{4}{{\sqrt 5 – 1}}\\ = \sqrt {14 + 6\sqrt 5 } – \dfrac{{4\left( {\sqrt 5 + 1} \right)}}{{\left( {\sqrt 5 – 1} \right)\left( {\sqrt 5 + 1} \right)}}\\ = \sqrt {{3^2} + 2.3.\sqrt 5 + {{\left( {\sqrt 5 } \right)}^2}} – \dfrac{{4\left( {\sqrt 5 + 1} \right)}}{{5 – 1}}\\ = \sqrt {{{\left( {3 + \sqrt 5 } \right)}^2}} – \left( {\sqrt 5 + 1} \right)\\ = \left| {3 + \sqrt 5 } \right| – \sqrt 5 + 1\\ = 3 + \sqrt 5 – \sqrt 5 + 1 = 4.\end{array}\) \(\begin{array}{l}d)\;\;\sqrt {27b} + 12\sqrt {\dfrac{1}{3}b} – \sqrt {48b} \left( {b > 0} \right)\\ = \sqrt {{3^2}.3b} + 12.\dfrac{{\sqrt {3b} }}{3} – \sqrt {{4^2}.3b} \\ = 3\sqrt {3b} + 4\sqrt {3b} – 4\sqrt {3b} = 3\sqrt {3b} .\end{array}\)
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