Bài 8 trang 29 Tài liệu dạy – học Toán 9 tập 1

Đề bài

Tính :

a) \(\dfrac{1}{{2 – \sqrt 5 }} + \dfrac{1}{{2 + \sqrt 5 }}\);

b) \(\dfrac{3}{2}\sqrt 6  + 2\sqrt {\dfrac{2}{3}}  – 4\sqrt {\dfrac{3}{2}} \);                                         

c) \(\dfrac{{2 + \sqrt 3 }}{{2 – \sqrt 3 }} – \dfrac{{2 – \sqrt 3 }}{{2 + \sqrt 3 }}\).   

d) \(\left( {2 + \dfrac{{\sqrt 5  – 5}}{{1 – \sqrt 5 }}} \right)\left( {2 – \dfrac{{\sqrt 5  + 5}}{{\sqrt 5  + 1}}} \right)\)

e) \(\dfrac{2}{{\sqrt 3  – 1}} + \dfrac{3}{{\sqrt 3  – 2}} + \dfrac{{12}}{{3 – \sqrt 3 }}\)

Lời giải chi tiết

\(a)\;\dfrac{1}{{2 – \sqrt 5 }} + \dfrac{1}{{2 + \sqrt 5 }} \)

\(= \dfrac{{2 + \sqrt 5  + 2 – \sqrt 5 }}{{\left( {2 – \sqrt 5 } \right)\left( {2 + \sqrt 5 } \right)}} \)

\(= \dfrac{4}{{4 – 5}} =  – 4.\)

\(\begin{array}{l}b)\;\;\dfrac{3}{2}\sqrt 6  + 2\sqrt {\dfrac{2}{3}}  – 4\sqrt {\dfrac{3}{2}}\\  = \dfrac{{3\sqrt 6 }}{2} + \dfrac{{2\sqrt {2.3} }}{3} – \dfrac{{4\sqrt {2.3} }}{2}\\ = \dfrac{{3\sqrt 6 }}{2} + \dfrac{{2\sqrt 6 }}{3} – 2\sqrt 6  = \dfrac{{\sqrt 6 }}{6}.\end{array}\)

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

\(\begin{array}{l}d)\;\;\left( {2 + \dfrac{{\sqrt 5  – 5}}{{1 – \sqrt 5 }}} \right)\left( {2 – \dfrac{{\sqrt 5  + 5}}{{\sqrt 5  + 1}}} \right)\\ = \left( {2 + \dfrac{{\sqrt 5 \left( {1 – \sqrt 5 } \right)}}{{1 – \sqrt 5 }}} \right)\left( {2 – \dfrac{{\sqrt 5 \left( {\sqrt 5  + 1} \right)}}{{\sqrt 5  + 1}}} \right)\\ = \left( {2 + \sqrt 5 } \right)\left( {2 – \sqrt 5 } \right) = 4 – 5 =  – 1.\end{array}\)

\(\begin{array}{l}e)\frac{2}{{\sqrt 3  – 1}} + \frac{3}{{\sqrt 3  – 2}} + \frac{{12}}{{3 – \sqrt 3 }}\\ = \frac{{2\left( {\sqrt 3  + 1} \right)}}{{3 – 1}} + \frac{{3\left( {\sqrt 3  + 2} \right)}}{{3 – 4}} + \frac{{12\left( {3 + \sqrt 3 } \right)}}{{9 – 3}}\\ = \frac{{2\left( {\sqrt 3  + 1} \right)}}{2} + \frac{{3\left( {\sqrt 3  + 2} \right)}}{{ – 1}} + \frac{{12\left( {3 + \sqrt 3 } \right)}}{6}\\ = \sqrt 3  + 1 – 3\sqrt 3  – 6 + 6 + 2\sqrt 3 \\ = 1\end{array}\)

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