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Trục căn thức ở mẫu và rút gọn ( nếu được):
a) \({{\sqrt 5 – \sqrt 3 } \over {\sqrt 2 }}\);
b) \({{26} \over {5 – 2\sqrt 3 }}\);
c) \({{2\sqrt {10} – 5} \over {4 – \sqrt {10} }}\);
d) \({{9 – 2\sqrt 3 } \over {3\sqrt 6 – 2\sqrt 2 }}\).
Gợi ý làm bài
a) \({{\sqrt 5 – \sqrt 3 } \over {\sqrt 2 }}\) \( = {{(\sqrt 5 – \sqrt 3 )\sqrt 2 } \over {{{(\sqrt 2 )}^2}}} = {{\sqrt {10} – \sqrt 6 } \over 2}\)
b) \({{26} \over {5 – 2\sqrt 3 }}\) \( = {{26(5 + 2\sqrt 3 )} \over {(5 – 2\sqrt 3 )(5 + 2\sqrt 3 )}} = {{26(5 + 2\sqrt 3 )} \over {25 – 12}}\)
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\( = {{26(5 + 2\sqrt 3 )} \over {13}} = 2(5 + 2\sqrt 3 ) = 10 + 4\sqrt 3 \)
c) \({{2\sqrt {10} – 5} \over {4 – \sqrt {10} }}\) \( = {{2\sqrt {2.5} – \sqrt {{5^2}} } \over {2\sqrt {{2^2}} – \sqrt {2.5} }}\)
\( = {{\sqrt 5 (2\sqrt 2 – \sqrt 5 )} \over {\sqrt 2 (2\sqrt 2 – \sqrt 5 )}} = {{\sqrt 5 } \over {\sqrt 2 }} = {{\sqrt 5 .\sqrt 2 } \over {{{(\sqrt 2 )}^2}}}\) \( = {{\sqrt {10} } \over 2}\)
d) \({{9 – 2\sqrt 3 } \over {3\sqrt 6 – 2\sqrt 2 }}\) \(= {{3\sqrt {{3^2}} – 2\sqrt 3 } \over {3\sqrt {3.2} – 2\sqrt 2 }}\)
\( = {{\sqrt 3 (3\sqrt 3 – 2)} \over {\sqrt 2 (3\sqrt 3 – 2)}} = {{\sqrt 3 } \over {\sqrt 2 }} = {{\sqrt {3.} \sqrt 2 } \over {{{(\sqrt 2 )}^2}}} = {{\sqrt 6 } \over 2}\)