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Rút gọn :
a) \(\left( {\dfrac{{\sqrt {15} – \sqrt 5 }}{{1 – \sqrt 3 }} + \dfrac{{\sqrt {14} – \sqrt 7 }}{{1 – \sqrt 2 }}} \right):\dfrac{1}{{\sqrt 7 – \sqrt 5 }}\);
b) \(\dfrac{{2\sqrt 5 – 5\sqrt 2 }}{{\sqrt 2 – \sqrt 5 }} + \dfrac{6}{{2 – \sqrt {10} }} + \sqrt {67 + 12\sqrt 7 } \);
c) \(\left( {\dfrac{{\sqrt 5 }}{{\sqrt 2 + 1}} + \dfrac{{14}}{{2\sqrt 2 – 1}} – \dfrac{6}{{2 – \sqrt 2 }}} \right).\sqrt {17 – 12\sqrt 2 } \).
+) Sử dụng công thức trục căn thức ở mẫu:\(\sqrt {\dfrac{A}{B}} = \sqrt {\dfrac{{A.B}}{{{B^2}}}} = \dfrac{{\sqrt {AB} }}{B},\;\;A\sqrt {\dfrac{B}{A}} = \sqrt {\dfrac{{{A^2}.B}}{A}} = \sqrt {AB} .\)
+) \(\dfrac{C}{{\sqrt A \pm B}} = \dfrac{{C\left( {\sqrt A \mp B} \right)}}{{A – {B^2}}};\;\;\dfrac{C}{{\sqrt A \pm \sqrt B }} = \dfrac{{C\left( {\sqrt A \mp \sqrt B } \right)}}{{A – B}}.\)
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+) \(\sqrt {{A^2}B} = \left| A \right|\sqrt B = \left\{ \begin{array}{l}A\sqrt B \;\;khi\;\;A \ge 0\\ – A\sqrt B \;\;khi\;\;A < 0\end{array} \right..\)
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\(\begin{array}{l}a)\;\left( {\dfrac{{\sqrt {15} – \sqrt 5 }}{{1 – \sqrt 3 }} + \dfrac{{\sqrt {14} – \sqrt 7 }}{{1 – \sqrt 2 }}} \right):\dfrac{1}{{\sqrt 7 – \sqrt 5 }}\\ = \left( {\dfrac{{\sqrt 5 \left( {\sqrt 3 – 1} \right)}}{{1 – \sqrt 3 }} + \dfrac{{\sqrt 7 \left( {\sqrt 2 – 1} \right)}}{{\left( {1 – \sqrt 2 } \right)}}} \right).\left( {\sqrt 7 – \sqrt 5 } \right)\\ = \left( { – \sqrt 5 – \sqrt 7 } \right)\left( {\sqrt 7 – \sqrt 5 } \right)\\ = – \left( {\sqrt 7 + \sqrt 5 } \right)\left( {\sqrt 7 – \sqrt 5 } \right)\\ = – \left( {7 – 5} \right) = – 2.\end{array}\) |
\(\begin{array}{l}b)\;\dfrac{{2\sqrt 5 – 5\sqrt 2 }}{{\sqrt 2 – \sqrt 5 }} + \dfrac{6}{{2 – \sqrt {10} }} + \sqrt {67 + 12\sqrt 7 } \\ = \dfrac{{\sqrt {10} \left( {\sqrt 2 – \sqrt 5 } \right)}}{{\sqrt 2 – \sqrt 5 }} + \dfrac{{6\left( {2 + \sqrt {10} } \right)}}{{4 – 10}} + \sqrt {{{\left( {3\sqrt 7 } \right)}^2} + 2.3\sqrt 7 .2 + {2^2}} \\ = \sqrt {10} – \left( {2 + \sqrt {10} } \right) + \sqrt {{{\left( {3\sqrt 7 + 2} \right)}^2}} \\ = \sqrt {10} – 2 – \sqrt {10} + \left| {3\sqrt 7 + 2} \right|\\ = – 2 + 3\sqrt 7 + 2 = 3\sqrt 7 .\end{array}\) |
\(\begin{array}{l}c)\;\left( {\dfrac{{\sqrt 5 }}{{\sqrt 2 + 1}} + \dfrac{{14}}{{2\sqrt 2 – 1}} – \dfrac{6}{{2 – \sqrt 2 }}} \right).\sqrt {17 – 12\sqrt 2 } \\ = \left[ {\dfrac{{\sqrt 5 \left( {\sqrt 2 – 1} \right)}}{{2 – 1}} + \dfrac{{14\left( {2\sqrt 2 + 1} \right)}}{{{{\left( {2\sqrt 2 } \right)}^2} – 1}} – \dfrac{{6\left( {2 + \sqrt 2 } \right)}}{{4 – {{\left( {\sqrt 2 } \right)}^2}}}} \right].\sqrt {{3^2} – 2.2\sqrt 2 .3 + {{\left( {2\sqrt 2 } \right)}^2}} \\ = \left( {\sqrt 5 \left( {\sqrt 2 – 1} \right) + \dfrac{{14\left( {2\sqrt 2 + 1} \right)}}{7} – \dfrac{{6\left( {2 + \sqrt 2 } \right)}}{2}} \right).\sqrt {{{\left( {3 – 2\sqrt 2 } \right)}^2}} \\ = \left[ {\sqrt 5 \left( {\sqrt 2 – 1} \right) + 2\left( {2\sqrt 2 + 1} \right) – 3\left( {2 + \sqrt 2 } \right)} \right].\left| {3 – 2\sqrt 2 } \right|\\ = \left[ {\sqrt 5 \left( {\sqrt 2 – 1} \right) + 4\sqrt 2 + 2 – 6 – 3\sqrt 2 } \right].\left( {3 – 2\sqrt 2 } \right)\\ = \left[ {\sqrt 5 \left( {\sqrt 2 – 1} \right) + \sqrt 2 – 4} \right].\left[ {{{\left( {\sqrt 2 } \right)}^2} – 2\sqrt 2 + 1} \right]\\ = \left[ {\sqrt 5 \left( {\sqrt 2 – 1} \right) – \sqrt 2 \left( {2\sqrt 2 – 1} \right)} \right]{\left( {\sqrt 2 – 1} \right)^2}\\ = \left[ {\sqrt 5 \left( {\sqrt 2 – 1} \right) – \sqrt 2 \left( {\sqrt 2 – 1} \right)\left( {2 + \sqrt 2 + 1} \right)} \right]{\left( {\sqrt 2 – 1} \right)^2}\\ = \left( {\sqrt 2 – 1} \right)\left( {\sqrt 5 – 2\sqrt 2 – 2 – \sqrt 2 } \right){\left( {\sqrt 2 – 1} \right)^2}\\ = {\left( {\sqrt 2 – 1} \right)^3}\left( {\sqrt 5 – 3\sqrt 2 – 2} \right).\end{array}\)