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Tính :
a) \(\dfrac{1}{{2 – \sqrt 5 }} + \dfrac{1}{{2 + \sqrt 5 }}\);
b) \(\dfrac{3}{2}\sqrt 6 + 2\sqrt {\dfrac{2}{3}} – 4\sqrt {\dfrac{3}{2}} \);
c) \(\dfrac{{2 + \sqrt 3 }}{{2 – \sqrt 3 }} – \dfrac{{2 – \sqrt 3 }}{{2 + \sqrt 3 }}\).
d) \(\left( {2 + \dfrac{{\sqrt 5 – 5}}{{1 – \sqrt 5 }}} \right)\left( {2 – \dfrac{{\sqrt 5 + 5}}{{\sqrt 5 + 1}}} \right)\)
e) \(\dfrac{2}{{\sqrt 3 – 1}} + \dfrac{3}{{\sqrt 3 – 2}} + \dfrac{{12}}{{3 – \sqrt 3 }}\)
+) 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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+) Quy đồng mẫu các phân thức rồi cộng các phân thức với nhau.
\(a)\;\dfrac{1}{{2 – \sqrt 5 }} + \dfrac{1}{{2 + \sqrt 5 }} = \dfrac{{2 + \sqrt 5 + 2 – \sqrt 5 }}{{\left( {2 – \sqrt 5 } \right)\left( {2 + \sqrt 5 } \right)}} = \dfrac{4}{{4 – 5}} = – 4.\)
\(\begin{array}{l}b)\;\;\dfrac{3}{2}\sqrt 6 + 2\sqrt {\dfrac{2}{3}} – 4\sqrt {\dfrac{3}{2}} = \dfrac{{3\sqrt 6 }}{2} + \dfrac{{2\sqrt {2.3} }}{3} – \dfrac{{4\sqrt {2.3} }}{2}\\ = \dfrac{{3\sqrt 6 }}{2} + \dfrac{{2\sqrt 6 }}{3} – 2\sqrt 6 = \dfrac{{\sqrt 6 }}{6}.\end{array}\)
\(\begin{array}{l}c)\;\dfrac{{2 + \sqrt 3 }}{{2 – \sqrt 3 }} – \dfrac{{2 – \sqrt 3 }}{{2 + \sqrt 3 }} = \dfrac{{{{\left( {2 + \sqrt 3 } \right)}^2} – {{\left( {2 – \sqrt 3 } \right)}^2}}}{{\left( {2 – \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)}}\\ = \dfrac{{4 + 4\sqrt 3 + 3 – \left( {4 – 4\sqrt 3 + 3} \right)}}{{4 – 3}} = \dfrac{{8\sqrt 3 }}{1} = 8\sqrt 3 .\end{array}\)
\(\begin{array}{l}d)\;\;\left( {2 + \dfrac{{\sqrt 5 – 5}}{{1 – \sqrt 5 }}} \right)\left( {2 – \dfrac{{\sqrt 5 + 5}}{{\sqrt 5 + 1}}} \right)\\ = \left( {2 + \dfrac{{\sqrt 5 \left( {1 – \sqrt 5 } \right)}}{{1 – \sqrt 5 }}} \right)\left( {2 – \dfrac{{\sqrt 5 \left( {\sqrt 5 + 1} \right)}}{{\sqrt 5 + 1}}} \right)\\ = \left( {2 + \sqrt 5 } \right)\left( {2 – \sqrt 5 } \right) = 4 – 5 = – 1.\end{array}\)
\(\begin{array}{l}e)\;\;\dfrac{2}{{\sqrt 3 – 1}} + \dfrac{3}{{\sqrt 3 – 2}} + \dfrac{{12}}{{3 – \sqrt 3 }}\\ = \dfrac{{3\left( {\sqrt 3 + 1} \right)}}{{3 – 1}} + \dfrac{{3\left( {\sqrt 3 + 2} \right)}}{{3 – 4}} + \dfrac{{12\left( {3 + \sqrt 3 } \right)}}{{9 – 3}}\\ = \dfrac{{3\sqrt 3 + 3}}{2} – 3\sqrt 3 – 6 + \dfrac{{12\left( {3 + \sqrt 3 } \right)}}{6}\\ = \dfrac{{3\sqrt 3 + 3}}{2} – 3\sqrt 3 – 6 + 6 + 2\sqrt 3 \\ = \dfrac{{3\sqrt 3 + 3}}{2} – 3\sqrt 3 \\ = \dfrac{{3\sqrt 3 + 3 – 6\sqrt 3 }}{2} = \dfrac{{3 – 3\sqrt 3 }}{2}.\end{array}\)