Bài 3 trang 31 Tài liệu dạy – học Toán 9 tập 1: Tính...

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: $\sqrt {{A^2}B}  = \left| A \right|\sqrt B  = \left\{ \begin{array}{l}A\;\;khi\;\;A \ge 0\\ - A\;\;\;khi\;\;A < 0\end{array} \right..$

$\begin{array}{l}a)\;\sqrt 2 \left( {\sqrt {4 - \sqrt 7 }  - \sqrt {4 + \sqrt 7 }  + \sqrt 2 } \right)\\ = \sqrt {8 - 2\sqrt 7 }  - \sqrt {8 + 2\sqrt 7 }  + 2\\ = \sqrt {{{\left( {\sqrt 7 } \right)}^2} - 2\sqrt 7  + 1}  - \sqrt {{{\left( {\sqrt 7 } \right)}^2} + 2\sqrt 7  + 1}  + 2\\ = \sqrt {{{\left( {\sqrt 7  - 1} \right)}^2}}  - \sqrt {{{\left( {\sqrt 7  + 1} \right)}^2}}  + 2\\ = \left| {\sqrt 7  - 1} \right| - \left| {\sqrt 7  + 1} \right| + 2\\ = \sqrt 7  - 1 - \sqrt 7  - 1 + 2 = 0\end{array}$

$\begin{array}{l}b)\;\left( {4 - \sqrt 7 } \right)\left( {\sqrt 2  + \sqrt {14} } \right)\sqrt {4 + \sqrt 7 } \\ = \dfrac{1}{2}.2\left( {4 - \sqrt 7 } \right).\sqrt 2 \left( {1 + \sqrt 7 } \right)\sqrt {4 + \sqrt 7 } \\ = \dfrac{1}{2}\left( {8 - 2\sqrt 7 } \right)\left( {1 + \sqrt 7 } \right)\sqrt {8 + 2\sqrt 7 } \\ = \dfrac{1}{2}\left( {{{\left( {\sqrt 7 } \right)}^2} - 2\sqrt 7  + 1} \right)\left( {1 + \sqrt 7 } \right)\sqrt {{{\left( {\sqrt 7 } \right)}^2} + 2\sqrt 7  + 1} \\ = \dfrac{1}{2}{\left( {\sqrt 7  - 1} \right)^2}\left( {1 + \sqrt 7 } \right)\left| {\sqrt 7  + 1} \right|\\ = \dfrac{1}{2}{\left( {\sqrt 7  - 1} \right)^2}\left( {1 + \sqrt 7 } \right)\left( {\sqrt 7  + 1} \right)\\ = \dfrac{1}{2}{\left( {\sqrt 7  - 1} \right)^2}{\left( {\sqrt 7  + 1} \right)^2}\\ = \dfrac{1}{2}{\left[ {\left( {\sqrt 7  - 1} \right)\left( {\sqrt 7  + 1} \right)} \right]^2}\\ = \dfrac{1}{2}{\left( {7 - 1} \right)^2} = \dfrac{1}{2}.36 = 18.\end{array}$

$\begin{array}{l}c)\;\sqrt {4 + \sqrt {15} }  - \sqrt {4 - \sqrt {15} }  - \sqrt {2 - \sqrt 3 } \\ = \dfrac{1}{2}.2\left( {\sqrt {4 + \sqrt {15} }  - \sqrt {4 - \sqrt {15} }  - \sqrt {2 - \sqrt 3 } } \right)\\ = \dfrac{1}{2}\left( {\sqrt {8 + 2\sqrt {15} }  - \sqrt {8 - 2\sqrt {15} }  - \sqrt {4 - 2\sqrt 3 } } \right)\\ = \dfrac{1}{2}\left( {\sqrt {{{\left( {\sqrt 5 } \right)}^2} + 2.\sqrt 5 .\sqrt 3  + {{\left( {\sqrt 3 } \right)}^2}}  - \sqrt {{{\left( {\sqrt 5 } \right)}^2} - 2.\sqrt 5 .\sqrt 3  + {{\left( {\sqrt 3 } \right)}^2}}  - \sqrt {{{\left( {\sqrt 3 } \right)}^2} + 2.\sqrt 3  + 1} } \right)\\ = \dfrac{1}{2}\left( {\sqrt {{{\left( {\sqrt 5  + \sqrt 3 } \right)}^2}}  - \sqrt {{{\left( {\sqrt 5  - \sqrt 3 } \right)}^2}}  - \sqrt {{{\left( {\sqrt 3  - 1} \right)}^2}} } \right)\\ = \dfrac{1}{2}\left( {\left| {\sqrt 5  + \sqrt 3 } \right| - \left| {\sqrt 5  - \sqrt 3 } \right| - \left| {\sqrt 3  - 1} \right|} \right)\\ = \dfrac{1}{2}\left( {\sqrt 5  + \sqrt 3  - \sqrt 5  + \sqrt 3  - \sqrt 3  - 1} \right)\\ = \dfrac{1}{2}.\left( {\sqrt 3  - 1} \right) = \dfrac{{\sqrt 3  - 1}}{2}.\end{array}$

$\begin{array}{l}d)\;\sqrt {13 + 30\sqrt {2 + \sqrt {9 + 4\sqrt 2 } } } \\ = \sqrt {13 + 30\sqrt {2 + \sqrt {{{\left( {2\sqrt 2 } \right)}^2} + 2.2\sqrt 2  + 1} } } \\ = \sqrt {13 + 30\sqrt {2 + \sqrt {{{\left( {2\sqrt 2  + 1} \right)}^2}} } } \\ = \sqrt {13 + 30\sqrt {2 + \left| {2\sqrt 2  + 1} \right|} } \\ = \sqrt {13 + 30\sqrt {2 + 2\sqrt 2  + 1} } \\ = \sqrt {13 + 30\sqrt {{{\left( {\sqrt 2  + 1} \right)}^2}} } \\ = \sqrt {13 + 30\left| {\sqrt 2  + 1} \right|} \\ = \sqrt {13 + 30\left( {\sqrt 2  + 1} \right)} \\ = \sqrt {13 + 30\sqrt 2  + 30} \\ = \sqrt {{{\left( {3\sqrt 2 } \right)}^2} + 2.3\sqrt 2 .5 + {5^2}} \\ = \sqrt {{{\left( {3\sqrt 2  + 5} \right)}^2}}  = \left| {3\sqrt 2  + 5} \right|\\ = 3\sqrt 2  + 5.\end{array}$