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Rút gọn :
a) \(\sqrt {{{\left( {\sqrt 7 – 2} \right)}^2}} – \sqrt {{{\left( {\sqrt 7 – 3} \right)}^2}} \);
b) \(\sqrt {7 – 2\sqrt {10} } – \sqrt {6 – 2\sqrt 5 } \);
c) \(\sqrt {42 – 10\sqrt {17} } + \sqrt {33 – 8\sqrt {17} } \);
d) \(\left( {2 + \sqrt 5 } \right)\sqrt {9 – 4\sqrt 5 } \);
e) \(\left( {3\sqrt 2 + \sqrt {10} } \right)\sqrt {28 – 12\sqrt 5 } \);
f) \(\sqrt {3 – \sqrt 5 } – \sqrt {3 + \sqrt 5 } \).
+) Sử dụng công thức: \(\sqrt {{A^2}B} = \left| A \right|\sqrt B = \left\{ \begin{array}{l}A\;\;khi\;\;A \ge 0\\ – A\;\;\;khi\;\;A < 0\end{array} \right..\)
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\(\begin{array}{l}a)\;\sqrt {{{\left( {\sqrt 7 – 2} \right)}^2}} – \sqrt {{{\left( {\sqrt 7 – 3} \right)}^2}} \\ = \left| {\sqrt 7 – 2} \right| – \left| {\sqrt 7 – 3} \right|\\ = \sqrt 7 – 2 – \sqrt 7 + 3\\ = 3 – 2 = 1.\end{array}\) \(\begin{array}{l}c)\;\sqrt {42 – 10\sqrt {17} } + \sqrt {33 – 8\sqrt {17} } \\ = \sqrt {{5^2} – 2.5.\sqrt {17} + 17} + \sqrt {17 – 2.4.\sqrt {17} + {4^2}} \\ = \sqrt {{{\left( {5 – \sqrt {17} } \right)}^2}} + \sqrt {{{\left( {\sqrt {17} – 4} \right)}^2}} \\ = \left| {5 – \sqrt {17} } \right| + \left| {\sqrt {17} – 4} \right|\\ = 5 – \sqrt {17} + \sqrt {17} – 4\\ = 5 – 4 = 1.\end{array}\) \(\begin{array}{l}e)\;\left( {3\sqrt 2 + \sqrt {10} } \right)\sqrt {28 – 12\sqrt 5 } \\ = \sqrt 2 \left( {3 + \sqrt 5 } \right)\sqrt 2 \sqrt {14 – 6\sqrt 5 } \\ = 2\left( {3 + \sqrt 5 } \right)\sqrt {{3^2} – 2.3\sqrt 5 + {{\left( {\sqrt 5 } \right)}^2}} \\ = 2\left( {3 + \sqrt 5 } \right)\sqrt {{{\left( {3 – \sqrt 5 } \right)}^2}} \\ = 2\left( {3 + \sqrt 5 } \right)\left| {3 – \sqrt 5 } \right|\\ = 2\left( {3 + \sqrt 5 } \right)\left( {3 – \sqrt 5 } \right)\\ = 2\left( {{3^2} – 5} \right)\\ = 2.4 = 8.\end{array}\) |
\(\begin{array}{l}b)\;\sqrt {7 – 2\sqrt {10} } – \sqrt {6 – 2\sqrt 5 } \\ = \sqrt {{{\left( {\sqrt 5 } \right)}^2} – 2.\sqrt 5 .\sqrt 2 + {{\left( {\sqrt 2 } \right)}^2}} – \sqrt {{{\left( {\sqrt 5 } \right)}^2} – 2\sqrt 5 + 1} \\ = \sqrt {{{\left( {\sqrt 5 – \sqrt 2 } \right)}^2}} – \sqrt {{{\left( {\sqrt 5 – 1} \right)}^2}} \\ = \left| {\sqrt 5 – \sqrt 2 } \right| – \left| {\sqrt 5 – 1} \right|\\ = \sqrt 5 – \sqrt 2 – \sqrt 5 + 1\\ = 1 – \sqrt 2 .\end{array}\) \(\begin{array}{l}d)\;\left( {2 + \sqrt 5 } \right)\sqrt {9 – 4\sqrt 5 } \\ = \left( {2 + \sqrt 5 } \right)\sqrt {{{\left( {\sqrt 5 } \right)}^2} – 2.2\sqrt 5 + 4} \\ = \left( {2 + \sqrt 5 } \right)\sqrt {{{\left( {\sqrt 5 – 2} \right)}^2}} \\ = \left( {2 + \sqrt 5 } \right)\left| {\sqrt 5 – 2} \right|\\ = \left( {2 + \sqrt 5 } \right)\left( {\sqrt 5 – 2} \right)\\ = {\left( {\sqrt 5 } \right)^2} – {2^2} = 5 – 4 = 1.\end{array}\) \(\begin{array}{l}f)\;\sqrt {3 – \sqrt 5 } – \sqrt {3 + \sqrt 5 } \\ = \dfrac{1}{{\sqrt 2 }}\left( {\sqrt {6 – 2\sqrt 5 } – \sqrt {6 + 2\sqrt 5 } } \right)\\ = \dfrac{1}{{\sqrt 2 }}\left( {\sqrt {{{\left( {\sqrt 5 } \right)}^2} – 2\sqrt 5 + 1} – \sqrt {{{\left( {\sqrt 5 } \right)}^2} + 2\sqrt 5 + 1} } \right)\\ = \dfrac{1}{{\sqrt 2 }}\left( {\sqrt {{{\left( {\sqrt 5 – 1} \right)}^2}} – \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} } \right)\\ = \dfrac{{\sqrt 2 }}{2}\left( {\left| {\sqrt 5 – 1} \right| – \left| {\sqrt 5 + 1} \right|} \right)\\ = \dfrac{{\sqrt 2 }}{2}\left( {\sqrt 5 – 1 – \sqrt 5 – 1} \right)\\ = \dfrac{{\sqrt 2 }}{2}\left( { – 2} \right) = – \sqrt 2 .\end{array}\) |