Rút gọn :
a) \(\dfrac{{3 + 2\sqrt 3 }}{{\sqrt 3 }} - \dfrac{{2\sqrt 3 + \sqrt {15} }}{{\sqrt 5 + 2}}\);
b) \(\dfrac{{5\sqrt 2 - 2\sqrt 5 }}{{\sqrt {10} }} - \dfrac{3}{{\sqrt 5 - \sqrt 2 }}\);
c) \(\dfrac{{\sqrt {15} - \sqrt {12} }}{{\sqrt 5 - 2}} - \dfrac{1}{{2 - \sqrt 3 }}\);
d) \(\dfrac{4}{{1 + \sqrt 3 }} + \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}\);
e) \(\sqrt {4 - 2\sqrt 3 } + \sqrt {\dfrac{2}{{2 - \sqrt 3 }}} - \sqrt {27} \);
f) \(\dfrac{{\sqrt 2 }}{{\sqrt {\sqrt 2 + 1} - 1}} - \dfrac{{\sqrt 2 }}{{\sqrt {\sqrt 2 + 1} + 1}}\).
+) 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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\(\begin{array}{l}a)\;\dfrac{{3 + 2\sqrt 3 }}{{\sqrt 3 }} - \dfrac{{2\sqrt 3 + \sqrt {15} }}{{\sqrt 5 + 2}}\\ = \dfrac{{\sqrt 3 \left( {\sqrt 3 + 2} \right)}}{{\sqrt 3 }} - \dfrac{{\sqrt 3 \left( {2 + \sqrt 5 } \right)}}{{\sqrt 5 + 2}}\\ = \sqrt 3 + 2 - \sqrt 3 = 2.\end{array}\) \(\begin{array}{l}c)\;\dfrac{{\sqrt {15} - \sqrt {12} }}{{\sqrt 5 - 2}} - \dfrac{1}{{2 - \sqrt 3 }}\\ = \dfrac{{\sqrt 3 \left( {\sqrt 5 - \sqrt 4 } \right)}}{{\sqrt 5 - 2}} - \dfrac{{2 + \sqrt 3 }}{{4 - 3}}\\ = \dfrac{{\sqrt 3 \left( {\sqrt 5 - 2} \right)}}{{\sqrt 5 - 2}} - \left( {2 + \sqrt 3 } \right)\\ = \sqrt 3 - 2 - \sqrt 3 = - 2.\end{array}\) \(\begin{array}{l}e)\;\sqrt {4 - 2\sqrt 3 } + \sqrt {\dfrac{2}{{2 - \sqrt 3 }}} - \sqrt {27} \\ = \sqrt {{{\left( {\sqrt 3 } \right)}^2} - 2\sqrt 3 + 1} + \sqrt {\dfrac{{2\left( {2 + \sqrt 3 } \right)}}{{4 - 3}}} - \sqrt {{3^2}.3} \\ = \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} + \sqrt {4 + 2\sqrt 3 } - 3\sqrt 3 \\ = \left| {\sqrt 3 - 1} \right| + \sqrt {{{\left( {\sqrt 3 } \right)}^2} - 2\sqrt 3 + 1} - 3\sqrt 3 \\ = \sqrt 3 - 1 + \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} - 3\sqrt 3 \;\;\;\left( {do\;\;\sqrt 3 - 1 > 0} \right)\\ = \sqrt 3 + 1 + \sqrt 3 + 1 - 3\sqrt 3 \\ = 2 - \sqrt 3 .\end{array}\)
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\(\begin{array}{l}b)\;\dfrac{{5\sqrt 2 - 2\sqrt 5 }}{{\sqrt {10} }} - \dfrac{3}{{\sqrt 5 - \sqrt 2 }}\\ = \dfrac{{\sqrt {10} \left( {\sqrt 5 - \sqrt 2 } \right)}}{{\sqrt {10} }} - \dfrac{{3\left( {\sqrt 5 + \sqrt 2 } \right)}}{{5 - 2}}\\ = \sqrt 5 - \sqrt 2 - \left( {\sqrt 5 + \sqrt 2 } \right)\\ = - 2\sqrt 2 .\end{array}\) \(\begin{array}{l}d)\;\dfrac{4}{{1 + \sqrt 3 }} + \dfrac{{\sqrt 3 - 1}}{{\sqrt 3 + 1}}\\ = \dfrac{{4 + \sqrt 3 - 1}}{{\sqrt 3 + 1}} = \dfrac{{3 + \sqrt 3 }}{{\sqrt 3 + 1}}\\ = \dfrac{{\sqrt 3 \left( {\sqrt 3 + 1} \right)}}{{\sqrt 3 + 1}} = 3.\end{array}\) \(\begin{array}{l}f)\;\dfrac{{\sqrt 2 }}{{\sqrt {\sqrt 2 + 1} - 1}} - \dfrac{{\sqrt 2 }}{{\sqrt {\sqrt 2 + 1} + 1}}\\ = \dfrac{{\sqrt 2 \left( {\sqrt {\sqrt 2 + 1} + 1} \right) - \sqrt 2 \left( {\sqrt {\sqrt 2 + 1} - 1} \right)}}{{\left( {\sqrt {\sqrt 2 + 1} + 1} \right)\left( {\sqrt {\sqrt 2 + 1} - 1} \right)}}\\ = \dfrac{{\sqrt {2\sqrt 2 + 2} + \sqrt 2 - \sqrt {2\sqrt 2 + 1} + \sqrt 2 }}{{\left( {\sqrt 2 + 1} \right) - 1}}\\ = \dfrac{{2\sqrt 2 }}{{\sqrt 2 }} = 2.\end{array}\) |