Bài 10 trang 30 Tài liệu dạy – học Toán 9 tập 1

Đề bài

Rút gọn :

a) \(\dfrac{{3 + 2\sqrt 3 }}{{\sqrt 3 }} - \dfrac{{2\sqrt 3  + \sqrt {15} }}{{\sqrt 5  + 2}}\);

b) \(\dfrac{{5\sqrt 2  - 2\sqrt 5 }}{{\sqrt {10} }} - \dfrac{3}{{\sqrt 5  - \sqrt 2 }}\);

c) \(\dfrac{{\sqrt {15}  - \sqrt {12} }}{{\sqrt 5  - 2}} - \dfrac{1}{{2 - \sqrt 3 }}\); 

d) \(\dfrac{4}{{1 + \sqrt 3 }} + \dfrac{{\sqrt 3  - 1}}{{\sqrt 3  + 1}}\);

e) \(\sqrt {4 - 2\sqrt 3 }  + \sqrt {\dfrac{2}{{2 - \sqrt 3 }}}  - \sqrt {27} \); 

f) \(\dfrac{{\sqrt 2 }}{{\sqrt {\sqrt 2  + 1}  - 1}} - \dfrac{{\sqrt 2 }}{{\sqrt {\sqrt 2  + 1}  + 1}}\).

Lời giải chi tiết

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

\(\begin{array}{l}b)\;\dfrac{{5\sqrt 2  - 2\sqrt 5 }}{{\sqrt {10} }} - \dfrac{3}{{\sqrt 5  - \sqrt 2 }}\\ = \dfrac{{\sqrt {10} \left( {\sqrt 5  - \sqrt 2 } \right)}}{{\sqrt {10} }} - \dfrac{{3\left( {\sqrt 5  + \sqrt 2 } \right)}}{{5 - 2}}\\ = \sqrt 5  - \sqrt 2  - \left( {\sqrt 5  + \sqrt 2 } \right)\\ =  - 2\sqrt 2 .\end{array}\)

\(\begin{array}{l}c)\;\dfrac{{\sqrt {15}  - \sqrt {12} }}{{\sqrt 5  - 2}} - \dfrac{1}{{2 - \sqrt 3 }}\\ = \dfrac{{\sqrt 3 \left( {\sqrt 5  - \sqrt 4 } \right)}}{{\sqrt 5  - 2}} - \dfrac{{2 + \sqrt 3 }}{{4 - 3}}\\ = \dfrac{{\sqrt 3 \left( {\sqrt 5  - 2} \right)}}{{\sqrt 5  - 2}} - \left( {2 + \sqrt 3 } \right)\\ = \sqrt 3  - 2 - \sqrt 3  =  - 2.\end{array}\)

\(\begin{array}{l}d)\;\dfrac{4}{{1 + \sqrt 3 }} + \dfrac{{\sqrt 3  - 1}}{{\sqrt 3  + 1}}\\ = \dfrac{{4 + \sqrt 3  - 1}}{{\sqrt 3  + 1}} = \dfrac{{3 + \sqrt 3 }}{{\sqrt 3  + 1}}\\ = \dfrac{{\sqrt 3 \left( {\sqrt 3  + 1} \right)}}{{\sqrt 3  + 1}} = \sqrt 3.\end{array}\)

\(\begin{array}{l}e)\;\sqrt {4 - 2\sqrt 3 }  + \sqrt {\dfrac{2}{{2 - \sqrt 3 }}}  - \sqrt {27} \\ = \sqrt {{{\left( {\sqrt 3 } \right)}^2} - 2\sqrt 3  + 1}  + \sqrt {\dfrac{{2\left( {2 + \sqrt 3 } \right)}}{{4 - 3}}}  - \sqrt {{3^2}.3} \\ = \sqrt {{{\left( {\sqrt 3  - 1} \right)}^2}}  + \sqrt {4 + 2\sqrt 3 }  - 3\sqrt 3 \\ = \left| {\sqrt 3  - 1} \right| + \sqrt {{{\left( {\sqrt 3 } \right)}^2} + 2\sqrt 3  + 1}  - 3\sqrt 3 \\ = \sqrt 3  - 1 + \sqrt {{{\left( {\sqrt 3  + 1} \right)}^2}}  - 3\sqrt 3 \;\;\;\left( {do\;\;\sqrt 3  - 1 > 0} \right)\\ = \sqrt 3  + 1 + \sqrt 3  + 1 - 3\sqrt 3 \\ = 2 - \sqrt 3 .\end{array}\)

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