Bài 12 trang 30 Tài liệu dạy – học Toán 9 tập 1

Đề bài

Rút gọn :

a) $\left( {\dfrac{{\sqrt {15}  - \sqrt 5 }}{{1 - \sqrt 3 }} + \dfrac{{\sqrt {14}  - \sqrt 7 }}{{1 - \sqrt 2 }}} \right):\dfrac{1}{{\sqrt 7  - \sqrt 5 }}$; 

b) $\dfrac{{2\sqrt 5  - 5\sqrt 2 }}{{\sqrt 2  - \sqrt 5 }} + \dfrac{6}{{2 - \sqrt {10} }} + \sqrt {67 + 12\sqrt 7 }$;

c) $\left( {\dfrac{{\sqrt 5 }}{{\sqrt 2  + 1}} + \dfrac{{14}}{{2\sqrt 2  - 1}} - \dfrac{6}{{2 - \sqrt 2 }}} \right).\sqrt {17 - 12\sqrt 2 }$.

Lời giải chi tiết

$\begin{array}{l}a)\;\left( {\dfrac{{\sqrt {15}  - \sqrt 5 }}{{1 - \sqrt 3 }} + \dfrac{{\sqrt {14}  - \sqrt 7 }}{{1 - \sqrt 2 }}} \right):\dfrac{1}{{\sqrt 7  - \sqrt 5 }}\\ = \left( {\dfrac{{\sqrt 5 \left( {\sqrt 3  - 1} \right)}}{{1 - \sqrt 3 }} + \dfrac{{\sqrt 7 \left( {\sqrt 2  - 1} \right)}}{{\left( {1 - \sqrt 2 } \right)}}} \right).\left( {\sqrt 7  - \sqrt 5 } \right)\\ = \left( { - \sqrt 5  - \sqrt 7 } \right)\left( {\sqrt 7  - \sqrt 5 } \right)\\ =  - \left( {\sqrt 7  + \sqrt 5 } \right)\left( {\sqrt 7  - \sqrt 5 } \right)\\ =  - \left( {7 - 5} \right) =  - 2.\end{array}$

$\begin{array}{l}b)\;\dfrac{{2\sqrt 5  - 5\sqrt 2 }}{{\sqrt 2  - \sqrt 5 }} + \dfrac{6}{{2 - \sqrt {10} }} + \sqrt {67 + 12\sqrt 7 } \\ = \dfrac{{\sqrt {10} \left( {\sqrt 2  - \sqrt 5 } \right)}}{{\sqrt 2  - \sqrt 5 }} + \dfrac{{6\left( {2 + \sqrt {10} } \right)}}{{4 - 10}} + \sqrt {{{\left( {3\sqrt 7 } \right)}^2} + 2.3\sqrt 7 .2 + {2^2}} \\ = \sqrt {10}  - \left( {2 + \sqrt {10} } \right) + \sqrt {{{\left( {3\sqrt 7  + 2} \right)}^2}} \\ = \sqrt {10}  - 2 - \sqrt {10}  + \left| {3\sqrt 7  + 2} \right|\\ =  - 2 + 3\sqrt 7  + 2 = 3\sqrt 7 .\end{array}$

$\begin{array}{l}c)\;\left( {\dfrac{{\sqrt 5 }}{{\sqrt 2  + 1}} + \dfrac{{14}}{{2\sqrt 2  - 1}} - \dfrac{6}{{2 - \sqrt 2 }}} \right).\sqrt {17 - 12\sqrt 2 } \\ = \left[ {\dfrac{{\sqrt 5 \left( {\sqrt 2  - 1} \right)}}{{2 - 1}} + \dfrac{{14\left( {2\sqrt 2  + 1} \right)}}{{{{\left( {2\sqrt 2 } \right)}^2} - 1}} - \dfrac{{6\left( {2 + \sqrt 2 } \right)}}{{4 - {{\left( {\sqrt 2 } \right)}^2}}}} \right].\sqrt {{3^2} - 2.2\sqrt 2 .3 + {{\left( {2\sqrt 2 } \right)}^2}} \\ = \left( {\sqrt 5 \left( {\sqrt 2  - 1} \right) + \dfrac{{14\left( {2\sqrt 2  + 1} \right)}}{7} - \dfrac{{6\left( {2 + \sqrt 2 } \right)}}{2}} \right).\sqrt {{{\left( {3 - 2\sqrt 2 } \right)}^2}} \\ = \left[ {\sqrt 5 \left( {\sqrt 2  - 1} \right) + 2\left( {2\sqrt 2  + 1} \right) - 3\left( {2 + \sqrt 2 } \right)} \right].\left| {3 - 2\sqrt 2 } \right|\\ = \left[ {\sqrt 5 \left( {\sqrt 2  - 1} \right) + 4\sqrt 2  + 2 - 6 - 3\sqrt 2 } \right].\left( {3 - 2\sqrt 2 } \right)\\ = \left[ {\sqrt 5 \left( {\sqrt 2  - 1} \right) + \sqrt 2  - 4} \right].\left[ {{{\left( {\sqrt 2 } \right)}^2} - 2\sqrt 2  + 1} \right]\\ = \left[ {\sqrt 5 \left( {\sqrt 2  - 1} \right) - \sqrt 2 \left( {2\sqrt 2  - 1} \right)} \right]{\left( {\sqrt 2  - 1} \right)^2}\\ = \left[ {\sqrt 5 \left( {\sqrt 2  - 1} \right) - \sqrt 2 \left( {\sqrt 2  - 1} \right)\left( {2 + \sqrt 2  + 1} \right)} \right]{\left( {\sqrt 2  - 1} \right)^2}\\ = \left( {\sqrt 2  - 1} \right)\left( {\sqrt 5  - 2\sqrt 2  - 2 - \sqrt 2 } \right){\left( {\sqrt 2  - 1} \right)^2}\\ = {\left( {\sqrt 2  - 1} \right)^3}\left( {\sqrt 5  - 3\sqrt 2  - 2} \right).\end{array}$

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