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Rút gọn các biểu thức:
a) \({2 \over {\sqrt 3 – 1}} – {2 \over {\sqrt 3 + 1}}\)
b) \({5 \over {12(2\sqrt 5 + 3\sqrt 2 )}} – {5 \over {12(2\sqrt 5 – 3\sqrt 2 )}}\)
c) \({{5 + \sqrt 5 } \over {5 – \sqrt 5 }} + {{5 – \sqrt 5 } \over {5 + \sqrt 5 }}\)
d) \({{\sqrt 3 } \over {\sqrt {\sqrt 3 + 1} – 1}} – {{\sqrt 3 } \over {\sqrt {\sqrt 3 + 1} + 1}}\)
Gợi ý làm bài
a) \({2 \over {\sqrt 3 – 1}} – {2 \over {\sqrt 3 + 1}}\) \(= {{2(\sqrt 3 + 1) – 2(\sqrt 3 – 1)} \over {(\sqrt 3 + 1)(\sqrt 3 – 1)}}\)
\( = {{2\sqrt 3 + 2 – 2\sqrt 3 + 2} \over {3 – 1}} = {4 \over 2} = 2\)
b) \({5 \over {12(2\sqrt 5 + 3\sqrt 2 )}} – {5 \over {12(2\sqrt 5 – 3\sqrt 2 )}}\)
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\( = {{5(2\sqrt 5 – 3\sqrt 2 ) – 5(2\sqrt 5 + 3\sqrt 2 )} \over {12(2\sqrt 5 + 3\sqrt 2 )(2\sqrt 5 – 3\sqrt 2 )}}\)
\(\eqalign{
& = {{10\sqrt 5 – 15\sqrt 2 – 10\sqrt 5 – 15\sqrt 2 } \over {12(20 – 18)}} \cr
& = {{ – 30\sqrt 2 } \over {12.2}} = – {{5\sqrt 2 } \over 4} \cr} \)
c) \({{5 + \sqrt 5 } \over {5 – \sqrt 5 }} + {{5 – \sqrt 5 } \over {5 + \sqrt 5 }}\) \(= {{{{(5 + \sqrt 5 )}^2} + {{(5 – \sqrt 5 )}^2}} \over {(5 + \sqrt 5 )(5 – \sqrt 5 )}}\)
\( = {{25 + 10\sqrt 5 + 5 + 25 – 10\sqrt 5 + 5} \over {25 – 5}} = {{60} \over {20}} = 3\)
d) \({{\sqrt 3 } \over {\sqrt {\sqrt 3 + 1} – 1}} – {{\sqrt 3 } \over {\sqrt {\sqrt 3 + 1} + 1}}\)
\( = {{\sqrt 3 (\sqrt {\sqrt 3 + 1} + 1) – \sqrt 3 (\sqrt {\sqrt 3 + 1} – 1)} \over {(\sqrt {\sqrt 3 + 1} + 1)(\sqrt {\sqrt 3 + 1} – 1)}}\)
\(\eqalign{
& = {{\sqrt {3(\sqrt 3 + 1)} + \sqrt 3 – \sqrt {3(\sqrt 3 + 1)} + \sqrt 3 } \over {\sqrt 3 + 1 – 1}} \cr
& = {{2\sqrt 3 } \over {\sqrt 3 }} = 2 \cr} \)