Bài 2 trang 39 Tài liệu dạy – học Toán 9 tập 1

Đề bài

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) \(\sqrt {3 - \sqrt 5 }  - \sqrt {3 + \sqrt 5 } \).

Lời giải chi tiết

\(\begin{array}{l}a)\;\sqrt {{{\left( {\sqrt 7  - 2} \right)}^2}}  - \sqrt {{{\left( {\sqrt 7  - 3} \right)}^2}} \\ = \left| {\sqrt 7  - 2} \right| - \left| {\sqrt 7  - 3} \right|\\ = \sqrt 7  - 2 - (3- \sqrt 7 ) (vì \sqrt 7  - 3 <0)\\=\sqrt 7 - 2 -3 +\sqrt 7=2.\sqrt 7 -5.\end{array}\)

\(\begin{array}{l}b)\;\sqrt {7 - 2\sqrt {10} }  - \sqrt {6 - 2\sqrt 5 } \\ = \sqrt {{{\left( {\sqrt 5 } \right)}^2} - 2.\sqrt 5 .\sqrt 2  + {{\left( {\sqrt 2 } \right)}^2}} \\\;\;\; - \sqrt {{{\left( {\sqrt 5 } \right)}^2} - 2\sqrt 5  + 1} \\ = \sqrt {{{\left( {\sqrt 5  - \sqrt 2 } \right)}^2}}  - \sqrt {{{\left( {\sqrt 5  - 1} \right)}^2}} \\ = \left| {\sqrt 5  - \sqrt 2 } \right| - \left| {\sqrt 5  - 1} \right|\\ = \sqrt 5  - \sqrt 2  - \sqrt 5  + 1\\ = 1 - \sqrt 2 .\end{array}\)

\(\begin{array}{l}c)\;\sqrt {42 - 10\sqrt {17} }  + \sqrt {33 - 8\sqrt {17} } \\ = \sqrt {{5^2} - 2.5.\sqrt {17}  + 17}  + \sqrt {17 - 2.4.\sqrt {17}  + {4^2}} \\ = \sqrt {{{\left( {5 - \sqrt {17} } \right)}^2}}  + \sqrt {{{\left( {\sqrt {17}  - 4} \right)}^2}} \\ = \left| {5 - \sqrt {17} } \right| + \left| {\sqrt {17}  - 4} \right|\\ = 5 - \sqrt {17}  + \sqrt {17}  - 4\\ = 5 - 4 = 1.\end{array}\)

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

\(\begin{array}{l}e)\;\left( {3\sqrt 2  + \sqrt {10} } \right)\sqrt {28 - 12\sqrt 5 } \\ = \sqrt 2 \left( {3 + \sqrt 5 } \right)\sqrt 2 \sqrt {14 - 6\sqrt 5 } \\ = 2\left( {3 + \sqrt 5 } \right)\sqrt {{3^2} - 2.3\sqrt 5  + {{\left( {\sqrt 5 } \right)}^2}} \\ = 2\left( {3 + \sqrt 5 } \right)\sqrt {{{\left( {3 - \sqrt 5 } \right)}^2}} \\ = 2\left( {3 + \sqrt 5 } \right)\left| {3 - \sqrt 5 } \right|\\ = 2\left( {3 + \sqrt 5 } \right)\left( {3 - \sqrt 5 } \right)\\ = 2\left( {{3^2} - 5} \right)\\ = 2.4 = 8.\end{array}\)

\(\begin{array}{l}f)\;\sqrt {3 - \sqrt 5 }  - \sqrt {3 + \sqrt 5 } \\ = \dfrac{1}{{\sqrt 2 }}\left( {\sqrt {6 - 2\sqrt 5 }  - \sqrt {6 + 2\sqrt 5 } } \right)\\ = \dfrac{1}{{\sqrt 2 }}\left( {\sqrt {{{\left( {\sqrt 5 } \right)}^2} - 2\sqrt 5  + 1}  - \sqrt {{{\left( {\sqrt 5 } \right)}^2} + 2\sqrt 5  + 1} } \right)\\ = \dfrac{1}{{\sqrt 2 }}\left( {\sqrt {{{\left( {\sqrt 5  - 1} \right)}^2}}  - \sqrt {{{\left( {\sqrt 5  + 1} \right)}^2}} } \right)\\ = \dfrac{{\sqrt 2 }}{2}\left( {\left| {\sqrt 5  - 1} \right| - \left| {\sqrt 5  + 1} \right|} \right)\\ = \dfrac{{\sqrt 2 }}{2}\left( {\sqrt 5  - 1 - \sqrt 5  - 1} \right)\\ = \dfrac{{\sqrt 2 }}{2}\left( { - 2} \right) =  - \sqrt 2 .\end{array}\)