Advertisements (Quảng cáo)
Tính :
a) \(\sqrt 2 \left( {\sqrt {4 – \sqrt 7 } – \sqrt {4 + \sqrt 7 } + \sqrt 2 } \right)\);
b) \(\left( {4 – \sqrt 7 } \right)\left( {\sqrt 2 + \sqrt {14} } \right)\sqrt {4 + \sqrt 7 } \);
c) \(\sqrt {4 + \sqrt {15} } – \sqrt {4 – \sqrt {15} } – \sqrt {2 – \sqrt 3 } \);
d) \(\sqrt {13 + 30\sqrt {2 + \sqrt {9 + 4\sqrt 2 } } } \).
+) 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)
\(\begin{array}{l}a)\;\sqrt 2 \left( {\sqrt {4 – \sqrt 7 } – \sqrt {4 + \sqrt 7 } + \sqrt 2 } \right)\\ = \sqrt {8 – 2\sqrt 7 } – \sqrt {8 + 2\sqrt 7 } + 2\\ = \sqrt {{{\left( {\sqrt 7 } \right)}^2} – 2\sqrt 7 + 1} – \sqrt {{{\left( {\sqrt 7 } \right)}^2} + 2\sqrt 7 + 1} + 2\\ = \sqrt {{{\left( {\sqrt 7 – 1} \right)}^2}} – \sqrt {{{\left( {\sqrt 7 + 1} \right)}^2}} + 2\\ = \left| {\sqrt 7 – 1} \right| – \left| {\sqrt 7 + 1} \right| + 2\\ = \sqrt 7 – 1 – \sqrt 7 – 1 + 2 = 0\end{array}\)
\(\begin{array}{l}b)\;\left( {4 – \sqrt 7 } \right)\left( {\sqrt 2 + \sqrt {14} } \right)\sqrt {4 + \sqrt 7 } \\ = \dfrac{1}{2}.2\left( {4 – \sqrt 7 } \right).\sqrt 2 \left( {1 + \sqrt 7 } \right)\sqrt {4 + \sqrt 7 } \\ = \dfrac{1}{2}\left( {8 – 2\sqrt 7 } \right)\left( {1 + \sqrt 7 } \right)\sqrt {8 + 2\sqrt 7 } \\ = \dfrac{1}{2}\left( {{{\left( {\sqrt 7 } \right)}^2} – 2\sqrt 7 + 1} \right)\left( {1 + \sqrt 7 } \right)\sqrt {{{\left( {\sqrt 7 } \right)}^2} + 2\sqrt 7 + 1} \\ = \dfrac{1}{2}{\left( {\sqrt 7 – 1} \right)^2}\left( {1 + \sqrt 7 } \right)\left| {\sqrt 7 + 1} \right|\\ = \dfrac{1}{2}{\left( {\sqrt 7 – 1} \right)^2}\left( {1 + \sqrt 7 } \right)\left( {\sqrt 7 + 1} \right)\\ = \dfrac{1}{2}{\left( {\sqrt 7 – 1} \right)^2}{\left( {\sqrt 7 + 1} \right)^2}\\ = \dfrac{1}{2}{\left[ {\left( {\sqrt 7 – 1} \right)\left( {\sqrt 7 + 1} \right)} \right]^2}\\ = \dfrac{1}{2}{\left( {7 – 1} \right)^2} = \dfrac{1}{2}.36 = 18.\end{array}\)
\(\begin{array}{l}c)\;\sqrt {4 + \sqrt {15} } – \sqrt {4 – \sqrt {15} } – \sqrt {2 – \sqrt 3 } \\ = \dfrac{1}{2}.2\left( {\sqrt {4 + \sqrt {15} } – \sqrt {4 – \sqrt {15} } – \sqrt {2 – \sqrt 3 } } \right)\\ = \dfrac{1}{2}\left( {\sqrt {8 + 2\sqrt {15} } – \sqrt {8 – 2\sqrt {15} } – \sqrt {4 – 2\sqrt 3 } } \right)\\ = \dfrac{1}{2}\left( {\sqrt {{{\left( {\sqrt 5 } \right)}^2} + 2.\sqrt 5 .\sqrt 3 + {{\left( {\sqrt 3 } \right)}^2}} – \sqrt {{{\left( {\sqrt 5 } \right)}^2} – 2.\sqrt 5 .\sqrt 3 + {{\left( {\sqrt 3 } \right)}^2}} – \sqrt {{{\left( {\sqrt 3 } \right)}^2} + 2.\sqrt 3 + 1} } \right)\\ = \dfrac{1}{2}\left( {\sqrt {{{\left( {\sqrt 5 + \sqrt 3 } \right)}^2}} – \sqrt {{{\left( {\sqrt 5 – \sqrt 3 } \right)}^2}} – \sqrt {{{\left( {\sqrt 3 – 1} \right)}^2}} } \right)\\ = \dfrac{1}{2}\left( {\left| {\sqrt 5 + \sqrt 3 } \right| – \left| {\sqrt 5 – \sqrt 3 } \right| – \left| {\sqrt 3 – 1} \right|} \right)\\ = \dfrac{1}{2}\left( {\sqrt 5 + \sqrt 3 – \sqrt 5 + \sqrt 3 – \sqrt 3 – 1} \right)\\ = \dfrac{1}{2}.\left( {\sqrt 3 – 1} \right) = \dfrac{{\sqrt 3 – 1}}{2}.\end{array}\)
\(\begin{array}{l}d)\;\sqrt {13 + 30\sqrt {2 + \sqrt {9 + 4\sqrt 2 } } } \\ = \sqrt {13 + 30\sqrt {2 + \sqrt {{{\left( {2\sqrt 2 } \right)}^2} + 2.2\sqrt 2 + 1} } } \\ = \sqrt {13 + 30\sqrt {2 + \sqrt {{{\left( {2\sqrt 2 + 1} \right)}^2}} } } \\ = \sqrt {13 + 30\sqrt {2 + \left| {2\sqrt 2 + 1} \right|} } \\ = \sqrt {13 + 30\sqrt {2 + 2\sqrt 2 + 1} } \\ = \sqrt {13 + 30\sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} } \\ = \sqrt {13 + 30\left| {\sqrt 2 + 1} \right|} \\ = \sqrt {13 + 30\left( {\sqrt 2 + 1} \right)} \\ = \sqrt {13 + 30\sqrt 2 + 30} \\ = \sqrt {{{\left( {3\sqrt 2 } \right)}^2} + 2.3\sqrt 2 .5 + {5^2}} \\ = \sqrt {{{\left( {3\sqrt 2 + 5} \right)}^2}} = \left| {3\sqrt 2 + 5} \right|\\ = 3\sqrt 2 + 5.\end{array}\)