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Tính :
a) \(\dfrac{{\sqrt {10} – \sqrt 2 }}{{\sqrt 5 – 1}} + \dfrac{{2 – \sqrt 2 }}{{1 – \sqrt 2 }}\);
b) \(\left( {\dfrac{{3\sqrt {125} }}{{15}} – \dfrac{{10 – 4\sqrt 5 }}{{\sqrt 5 – 2}}} \right)\dfrac{1}{{\sqrt 5 }}\);
c) \(\dfrac{2}{{\sqrt 7 + \sqrt 3 }} + \sqrt {\dfrac{2}{{5 – \sqrt {21} }}} \);
d) \(\sqrt {\dfrac{{\sqrt 3 }}{{8\sqrt 3 + 3\sqrt {21} }}} \left( {3\sqrt 2 + \sqrt {14} } \right)\).
+) Sử dụng công thức trục căn thức ở mẫu:\(\sqrt {\dfrac{A}{B}} = \sqrt {\dfrac{{A.B}}{{{B^2}}}} = \dfrac{{\sqrt {AB} }}{B},\;\;A\sqrt {\dfrac{B}{A}} = \sqrt {\dfrac{{{A^2}.B}}{A}} = \sqrt {AB} .\)
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+) \(\dfrac{C}{{\sqrt A \pm B}} = \dfrac{{C\left( {\sqrt A \mp B} \right)}}{{A – {B^2}}};\;\;\dfrac{C}{{\sqrt A \pm \sqrt B }} = \dfrac{{C\left( {\sqrt A \mp \sqrt B } \right)}}{{A – B}}.\)
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\(\begin{array}{l}a)\;\dfrac{{\sqrt {10} – \sqrt 2 }}{{\sqrt 5 – 1}} + \dfrac{{2 – \sqrt 2 }}{{1 – \sqrt 2 }}\\ = \dfrac{{\sqrt 2 \left( {\sqrt 5 – 1} \right)}}{{\sqrt 5 – 1}} + \dfrac{{\sqrt 2 \left( {\sqrt 2 – 1} \right)}}{{1 – \sqrt 2 }}\\ = \sqrt 2 – \sqrt 2 = 0.\end{array}\) \(\begin{array}{l}c)\;\dfrac{2}{{\sqrt 7 + \sqrt 3 }} + \sqrt {\dfrac{2}{{5 – \sqrt {21} }}} \\ = \dfrac{{2\left( {\sqrt 7 – \sqrt 3 } \right)}}{{{{\left( {\sqrt 7 } \right)}^2} – {{\left( {\sqrt 3 } \right)}^2}}} + \sqrt {\dfrac{{2\left( {5 + \sqrt {21} } \right)}}{{{5^2} – {{\left( {\sqrt {21} } \right)}^2}}}} \\ = \dfrac{{2\left( {\sqrt 7 – \sqrt 3 } \right)}}{{7 – 3}} + \sqrt {\dfrac{{10 + 2\sqrt {21} }}{4}} \\ = \dfrac{{2\left( {\sqrt 7 – \sqrt 3 } \right)}}{4} + \dfrac{{\sqrt {{{\left( {\sqrt 7 } \right)}^2} + 2\sqrt 7 .\sqrt 3 + {{\left( {\sqrt 3 } \right)}^2}} }}{2}\\ = \dfrac{{\sqrt 7 – \sqrt 3 }}{2} + \dfrac{{\sqrt {{{\left( {\sqrt 7 + \sqrt 3 } \right)}^2}} }}{2}\\ = \dfrac{{\sqrt 7 – \sqrt 3 }}{2} + \dfrac{{\left| {\sqrt 7 + \sqrt 3 } \right|}}{2}\\ = \dfrac{{\sqrt 7 – \sqrt 3 }}{2} + \dfrac{{\sqrt 7 + \sqrt 3 }}{2}\\ = \dfrac{{\sqrt 7 – \sqrt 3 + \sqrt 7 + \sqrt 3 }}{2}\\ = \dfrac{{2\sqrt 7 }}{2} = \sqrt 7 .\end{array}\) |
\(\begin{array}{l}b)\;\left( {\dfrac{{3\sqrt {125} }}{{15}} – \dfrac{{10 – 4\sqrt 5 }}{{\sqrt 5 – 2}}} \right)\dfrac{1}{{\sqrt 5 }}\\ = \left( {\dfrac{{3\sqrt {{5^2}.5} }}{{15}} – \dfrac{{2\sqrt 5 \left( {\sqrt 5 – 2} \right)}}{{\sqrt 5 – 2}}} \right).\dfrac{1}{{\sqrt 5 }}\\ = \left( {\dfrac{{3.5\sqrt 5 }}{{15}} – 2\sqrt 5 } \right).\dfrac{1}{{\sqrt 5 }}\\ = \left( {\sqrt 5 – 2\sqrt 5 } \right).\dfrac{1}{{\sqrt 5 }}\\ = – \sqrt 5 .\dfrac{1}{{\sqrt 5 }} = – 1.\end{array}\) \(\begin{array}{l}d)\;\sqrt {\dfrac{{\sqrt 3 }}{{8\sqrt 3 + 3\sqrt {21} }}} \left( {3\sqrt 2 + \sqrt {14} } \right)\\ = \sqrt {\dfrac{{\sqrt 3 }}{{\sqrt 3 \left( {8 + 3\sqrt 7 } \right)}}} .\sqrt 2 \left( {3 + \sqrt 7 } \right)\\ = \sqrt {\dfrac{1}{{8 + 3\sqrt 7 }}} .\sqrt 2 \left( {3 + \sqrt 7 } \right)\\ = \sqrt 2 \sqrt {\dfrac{{8 – 3\sqrt 7 }}{{{8^2} – {{\left( {3\sqrt 7 } \right)}^2}}}} .\left( {3 + \sqrt 7 } \right)\\ = \sqrt {\dfrac{{16 – 6\sqrt 7 }}{{64 – 63}}} .\left( {3 + \sqrt 7 } \right)\\ = \sqrt {16 – 6\sqrt 7 } \left( {3 + \sqrt 7 } \right)\\ = \sqrt {{3^2} – 2.3\sqrt 7 + {{\left( {\sqrt 7 } \right)}^2}} \left( {3 + \sqrt 7 } \right)\\ = \sqrt {{{\left( {3 – \sqrt 7 } \right)}^2}} \left( {3 + \sqrt 7 } \right)\\ = \left| {3 – \sqrt 7 } \right|\left( {3 + \sqrt 7 } \right)\\ = \left( {3 – \sqrt 7 } \right)\left( {3 + \sqrt 7 } \right)\\ = {3^2} – 7 = 9 – 7 = 2.\end{array}\) |