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a) \({1 \over {\sqrt 2 + \root 3 \of 3 }}\) b) \({1 \over {\sqrt 2 + \sqrt 3 + \sqrt 5 }}\)
Giải
a)
\({1 \over {\sqrt 2 + \root 3 \of 3 }} = {{\root 3 \of 3 – \sqrt 2 } \over {{{\left( {\root 3 \of 3 } \right)}^2} – {{\left( {\sqrt 2 } \right)}^2}}} = {{\root 3 \of 3 – \sqrt 2 } \over {\root 3 \of 9 – 2}}\)
\( = {{\left( {\root 3 \of 3 – \sqrt 2 } \right)\left( {\root 3 \of {{9^2}} + 2\root 3 \of 9 + 4} \right)} \over {{{\left( {\root 3 \of 9 } \right)}^3} – {2^3}}} = {{\left( {\root 3 \of 3 – \sqrt 2 } \right)\left( {3\root 3 \of {{3}} + 2\root 3 \of 9 + 4} \right)} \over 1}\)
b) \({1 \over {\sqrt 2 + \sqrt 3 + \sqrt 5 }} = {{\sqrt 2 + \sqrt 3 – \sqrt 5 } \over {{{\left( {\sqrt 2 + \sqrt 3 } \right)}^2} – 5}} = {{\sqrt 2 + \sqrt 3 – \sqrt 5 } \over {2\sqrt 6 }}\)
\(= {{\sqrt 6 \left( {\sqrt 2 + \sqrt 3 – \sqrt 5 } \right)} \over {12}}\)