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Trục căn thức ở mẫu:
a) \({1 \over {\sqrt 3 + \sqrt 2 + 1}}\)
b)\({1 \over {\sqrt 5 – \sqrt 3 + 2}}\)
Gợi ý làm bài
a) \(\eqalign{
& {1 \over {\sqrt 3 + \sqrt 2 + 1}} = {1 \over {\sqrt 3 + (\sqrt 2 + 1)}} \cr
& = {{\sqrt 3 – (\sqrt 2 + 1)} \over {\left[ {\sqrt 3 + (\sqrt 2 + 1)} \right]\left[ {\sqrt 3 – (\sqrt 2 + 1)} \right]}} \cr} \)
\( = {{\sqrt 3 – \sqrt 2 – 1} \over {3 – {{(\sqrt 2 + 1)}^2}}} = {{\sqrt 3 – \sqrt 2 – 1} \over {3 – (2 + 2\sqrt 2 + 1)}} = {{\sqrt 3 – \sqrt 2 – 1} \over { – 2\sqrt 2 }}\)
\( = {{ – \sqrt 2 (\sqrt 3 – \sqrt 2 – 1)} \over {2{{(\sqrt 2 )}^2}}} = {{ – \sqrt 6 + 2 + \sqrt 2 } \over 4}\)
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b) \({1 \over {\sqrt 5 – \sqrt 3 + 2}} = {{\sqrt 5 + (\sqrt 3 – 2)} \over {\left[ {\sqrt 5 – (\sqrt 3 – 2)} \right]\left[ {\sqrt 5 + (\sqrt 3 – 2)} \right]}}\)
\( = {{\sqrt 5 + (\sqrt 3 – 2)} \over {5 – {{(\sqrt 3 – 2)}^2}}} = {{\sqrt 5 + (\sqrt 3 – 2)} \over {5 – (3 – 4\sqrt 3 + 4)}} = {{\sqrt 5 + (\sqrt 3 – 2)} \over {4\sqrt 3 – 2}}\)
\(= {{\sqrt 5 + \sqrt 3 – 2} \over {2(2\sqrt 3 – 1)}} = {{(\sqrt 5 + \sqrt 3 – 2)(2\sqrt 3 + 1)} \over {2\left[ {(2\sqrt 3 – 1)(2\sqrt 3 + 1)} \right]}}\)
\(\eqalign{
& = {{2\sqrt {15} + \sqrt 5 + 6 + \sqrt 3 – 4\sqrt 3 – 2} \over {2(12 – 1)}} \cr
& = {{2\sqrt {15} + \sqrt 5 + 4 – 3\sqrt 3 } \over {22}} \cr} \)