Bài 4 trang 31 Tài liệu dạy học Toán lớp 9 tập 1: Tính...

$\begin{array}{l}a)\;\dfrac{{\sqrt {10}  - \sqrt 2 }}{{\sqrt 5  - 1}} + \dfrac{{2 - \sqrt 2 }}{{1 - \sqrt 2 }}\\ = \dfrac{{\sqrt 2 \left( {\sqrt 5  - 1} \right)}}{{\sqrt 5  - 1}} + \dfrac{{\sqrt 2 \left( {\sqrt 2  - 1} \right)}}{{1 - \sqrt 2 }}\\ = \sqrt 2  - \sqrt 2  = 0.\end{array}$

$\begin{array}{l}c)\;\dfrac{2}{{\sqrt 7  + \sqrt 3 }} + \sqrt {\dfrac{2}{{5 - \sqrt {21} }}} \\ = \dfrac{{2\left( {\sqrt 7  - \sqrt 3 } \right)}}{{{{\left( {\sqrt 7 } \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}}} + \sqrt {\dfrac{{2\left( {5 + \sqrt {21} } \right)}}{{{5^2} - {{\left( {\sqrt {21} } \right)}^2}}}} \\ = \dfrac{{2\left( {\sqrt 7  - \sqrt 3 } \right)}}{{7 - 3}} + \sqrt {\dfrac{{10 + 2\sqrt {21} }}{4}} \\ = \dfrac{{2\left( {\sqrt 7  - \sqrt 3 } \right)}}{4} + \dfrac{{\sqrt {{{\left( {\sqrt 7 } \right)}^2} + 2\sqrt 7 .\sqrt 3  + {{\left( {\sqrt 3 } \right)}^2}} }}{2}\\ = \dfrac{{\sqrt 7  - \sqrt 3 }}{2} + \dfrac{{\sqrt {{{\left( {\sqrt 7  + \sqrt 3 } \right)}^2}} }}{2}\\ = \dfrac{{\sqrt 7  - \sqrt 3 }}{2} + \dfrac{{\left| {\sqrt 7  + \sqrt 3 } \right|}}{2}\\ = \dfrac{{\sqrt 7  - \sqrt 3 }}{2} + \dfrac{{\sqrt 7  + \sqrt 3 }}{2}\\ = \dfrac{{\sqrt 7  - \sqrt 3  + \sqrt 7  + \sqrt 3 }}{2}\\ = \dfrac{{2\sqrt 7 }}{2} = \sqrt 7 .\end{array}$

$\begin{array}{l}b)\;\left( {\dfrac{{3\sqrt {125} }}{{15}} - \dfrac{{10 - 4\sqrt 5 }}{{\sqrt 5  - 2}}} \right)\dfrac{1}{{\sqrt 5 }}\\ = \left( {\dfrac{{3\sqrt {{5^2}.5} }}{{15}} - \dfrac{{2\sqrt 5 \left( {\sqrt 5  - 2} \right)}}{{\sqrt 5  - 2}}} \right).\dfrac{1}{{\sqrt 5 }}\\ = \left( {\dfrac{{3.5\sqrt 5 }}{{15}} - 2\sqrt 5 } \right).\dfrac{1}{{\sqrt 5 }}\\ = \left( {\sqrt 5  - 2\sqrt 5 } \right).\dfrac{1}{{\sqrt 5 }}\\ =  - \sqrt 5 .\dfrac{1}{{\sqrt 5 }} =  - 1.\end{array}$

$\begin{array}{l}d)\;\sqrt {\dfrac{{\sqrt 3 }}{{8\sqrt 3  + 3\sqrt {21} }}} \left( {3\sqrt 2  + \sqrt {14} } \right)\\ = \sqrt {\dfrac{{\sqrt 3 }}{{\sqrt 3 \left( {8 + 3\sqrt 7 } \right)}}} .\sqrt 2 \left( {3 + \sqrt 7 } \right)\\ = \sqrt {\dfrac{1}{{8 + 3\sqrt 7 }}} .\sqrt 2 \left( {3 + \sqrt 7 } \right)\\ = \sqrt 2 \sqrt {\dfrac{{8 - 3\sqrt 7 }}{{{8^2} - {{\left( {3\sqrt 7 } \right)}^2}}}} .\left( {3 + \sqrt 7 } \right)\\ = \sqrt {\dfrac{{16 - 6\sqrt 7 }}{{64 - 63}}} .\left( {3 + \sqrt 7 } \right)\\ = \sqrt {16 - 6\sqrt 7 } \left( {3 + \sqrt 7 } \right)\\ = \sqrt {{3^2} - 2.3\sqrt 7  + {{\left( {\sqrt 7 } \right)}^2}} \left( {3 + \sqrt 7 } \right)\\ = \sqrt {{{\left( {3 - \sqrt 7 } \right)}^2}} \left( {3 + \sqrt 7 } \right)\\ = \left| {3 - \sqrt 7 } \right|\left( {3 + \sqrt 7 } \right)\\ = \left( {3 - \sqrt 7 } \right)\left( {3 + \sqrt 7 } \right)\\ = {3^2} - 7 = 9 - 7 = 2.\end{array}$