Advertisements (Quảng cáo)
Tính :
a) \(\sqrt {9 – 4\sqrt 5 } – \sqrt {14 + 6\sqrt 5 } \);
b) \(\left( {3\sqrt 2 + \sqrt {10} } \right)\sqrt {28 – 12\sqrt 5 } \);
c) \(\sqrt {13 – \sqrt {160} } – \sqrt {53 + 4\sqrt {60} } \);
d) \(\sqrt {2 + \sqrt 3 } \left( {\sqrt 6 – \sqrt 2 } \right)\).
+) Sử dụng công thức: \(\sqrt {{A^2}B} = \left| A \right|\sqrt B = \left\{ \begin{array}{l}A\;\;khi\;\;A \ge 0\\ – A\;\;\;khi\;\;A < 0\end{array} \right..\)
Advertisements (Quảng cáo)
|
\(\begin{array}{l}a)\;\sqrt {9 – 4\sqrt 5 } – \sqrt {14 + 6\sqrt 5 } \\ = \sqrt {{{\left( {\sqrt 5 } \right)}^2} – 2.2\sqrt 5 + {2^2}} – \sqrt {{3^2} + 2.3.\sqrt 5 + {{\left( {\sqrt 5 } \right)}^2}} \\ = \sqrt {{{\left( {\sqrt 5 – 2} \right)}^2}} – \sqrt {{{\left( {3 + \sqrt 5 } \right)}^2}} \\ = \left| {\sqrt 5 – 2} \right| – \left| {3 + \sqrt 5 } \right|\\ = \sqrt 5 – 2 – 3 – \sqrt 5 = – 5.\end{array}\) \(\begin{array}{l}c)\;\;\sqrt {13 – \sqrt {160} } – \sqrt {53 + 4\sqrt {60} } \\ = \sqrt {13 – \sqrt {{4^2}.10} } – \sqrt {53 + 4.\sqrt {{2^2}.15} } \\ = \sqrt {13 – 4\sqrt {10} } – \sqrt {53 + 8\sqrt {15} } \\ = \sqrt {{{\left( {2\sqrt 2 } \right)}^2} – 2.2\sqrt 2 .\sqrt 5 + {{\left( {\sqrt 5 } \right)}^2}} – \sqrt {{{\left( {4\sqrt 3 } \right)}^2} + 2.4\sqrt 3 .\sqrt 5 + {{\left( {\sqrt 5 } \right)}^2}} \\ = \sqrt {{{\left( {2\sqrt 2 – \sqrt 5 } \right)}^2}} – \sqrt {{{\left( {4\sqrt 3 + \sqrt 5 } \right)}^2}} \\ = \left| {2\sqrt 2 – \sqrt 5 } \right| – \left| {4\sqrt 3 + \sqrt 5 } \right|\\ = 2\sqrt 2 – \sqrt 5 – 4\sqrt 3 – \sqrt 5 \\ = 2\sqrt 2 – 2\sqrt 5 – 4\sqrt 3 .\end{array}\) |
\(\begin{array}{l}b)\;\;\left( {3\sqrt 2 + \sqrt {10} } \right)\sqrt {28 – 12\sqrt 5 } \\ = \sqrt 2 \left( {3 + \sqrt 5 } \right)\sqrt {28 – 12\sqrt 5 } \\ = \left( {3 + \sqrt 5 } \right)\sqrt {56 – 24\sqrt 5 } \\ = \left( {3 + \sqrt 5 } \right)\sqrt {{6^2} – 2.6.2\sqrt 5 + {{\left( {2\sqrt 5 } \right)}^2}} \\ = \left( {3 + \sqrt 5 } \right)\sqrt {{{\left( {6 – 2\sqrt 5 } \right)}^2}} \\ = \left( {3 + \sqrt 5 } \right)\left| {6 – 2\sqrt 5 } \right|\\ = \left( {3 + \sqrt 5 } \right).2\left( {3 – \sqrt 5 } \right)\\ = 2\left( {{3^2} – 5} \right) = 8.\end{array}\) \(\begin{array}{l}d)\;\;\sqrt {2 + \sqrt 3 } \left( {\sqrt 6 – \sqrt 2 } \right)\\ = \sqrt {2 + \sqrt 3 } .\sqrt 2 \left( {\sqrt 3 – 1} \right)\\ = \sqrt {4 + 2\sqrt 3 } \left( {\sqrt 3 – 1} \right)\\ = \sqrt {3 + 2\sqrt 3 + 1} \left( {\sqrt 3 – 1} \right)\\ = \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} .\left( {\sqrt 3 – 1} \right)\\ = \left| {\sqrt 3 + 1} \right|\left( {\sqrt 3 – 1} \right)\\ = \left( {\sqrt 3 + 1} \right)\left( {\sqrt 3 – 1} \right)\\ = 3 – 2 = 1.\end{array}\) |