Đề bài
Trục căn thức ở mẫu [ giả thiết các biểu thức chữ đều có nghĩa ] :
a] \[\dfrac{7}{{\sqrt 3 }};\;\dfrac{3}{{2\sqrt 5 }};\;\dfrac{5}{{3\sqrt {12} }};\;\dfrac{2}{{3\sqrt {20} }}\];
b] \[\dfrac{{\sqrt 3 + 3}}{{5\sqrt 3 }};\;\dfrac{{7 - \sqrt 7 }}{{\sqrt 7 - 1}}\];
c] \[\dfrac{5}{{\sqrt 5 + 2}};\;\dfrac{3}{{\;\sqrt 3 - 1}};\;\dfrac{2}{{\sqrt 5 + \sqrt 3 }};\]\[\;\dfrac{{\sqrt 5 + 2}}{{\sqrt 5 - 2}};\;\dfrac{{\sqrt 7 - \sqrt 5 }}{{\sqrt 7 + \sqrt 5 }}\];
d] \[\dfrac{{y + a\sqrt y }}{{a\sqrt y }};\;\dfrac{{b - \sqrt b }}{{\sqrt b - 1}}\];
e] \[\dfrac{b}{{5 + \sqrt b }};\;\dfrac{p}{{2\sqrt p - 1}};\;\dfrac{{\sqrt a + \sqrt b }}{{\sqrt a - \sqrt b }}\].
Phương pháp giải - Xem chi tiết
+] Sử dụng công thức trục căn thức ở mẫu:\[\sqrt {\dfrac{A}{B}} = \sqrt {\dfrac{{A.B}}{{{B^2}}}} = \dfrac{{\sqrt {AB} }}{B},\]\[\;\;A\sqrt {\dfrac{B}{A}} = \sqrt {\dfrac{{{A^2}.B}}{A}} = \sqrt {AB} .\]
+] \[\dfrac{C}{{\sqrt A \pm B}} = \dfrac{{C\left[ {\sqrt A \mp B} \right]}}{{A - {B^2}}};\]\[\;\;\dfrac{C}{{\sqrt A \pm \sqrt B }} = \dfrac{{C\left[ {\sqrt A \mp \sqrt B } \right]}}{{A - B}}.\]
Lời giải chi tiết
\[\begin{array}{l}a]\;\;\dfrac{7}{{\sqrt 3 }} = \dfrac{{7\sqrt 3 }}{3};\\\dfrac{3}{{2\sqrt 5 }} = \dfrac{{3\sqrt 5 }}{{2.5}} = \dfrac{{3\sqrt 5 }}{{10}}\\\;\;\;\dfrac{5}{{3\sqrt {12} }} = \dfrac{5}{{3\sqrt {{2^2}.3} }} = \dfrac{5}{{6\sqrt 3 }} = \dfrac{{5\sqrt 3 }}{{18}}\\\;\;\;\dfrac{2}{{3\sqrt {20} }} = \dfrac{2}{{3\sqrt {{2^2}.5} }} = \dfrac{2}{{6\sqrt 5 }} = \dfrac{{2\sqrt 5 }}{{30}} = \dfrac{{\sqrt 5 }}{{15}}.\end{array}\]
\[\begin{array}{l}b]\;\dfrac{{\sqrt 3 + 3}}{{5\sqrt 3 }} = \dfrac{{\sqrt 3 \left[ {1 + \sqrt 3 } \right]}}{{5\sqrt 3 }} = \dfrac{{\sqrt 3 + 1}}{5}\\\;\;\;\dfrac{{7 - \sqrt 7 }}{{\sqrt 7 - 1}} = \dfrac{{\sqrt 7 \left[ {\sqrt 7 - 1} \right]}}{{\sqrt 7 - 1}} = \sqrt 7 .\end{array}\]
\[\begin{array}{l}c]\;\;\dfrac{5}{{\sqrt 5 + 2}} = \dfrac{{5\left[ {\sqrt 5 - 2} \right]}}{{5 - {2^2}}} \\\;\;\;= \dfrac{{5\sqrt 5 - 10}}{{5 - 4}} = 5\sqrt 5 - 10\\\;\;\;\;\;\dfrac{3}{{\sqrt 3 - 1}} = \dfrac{{3\left[ {\sqrt 3 + 1} \right]}}{{3 - 1}} = \dfrac{{3\sqrt 3 + 3}}{2}\\\;\;\;\;\;\dfrac{2}{{\sqrt 5 + \sqrt 3 }} = \dfrac{{2\left[ {\sqrt 5 - \sqrt 3 } \right]}}{{5 - 3}} \\\;\;\;= \dfrac{{2\left[ {\sqrt 5 - \sqrt 3 } \right]}}{2} = \sqrt 5 - \sqrt 3 \\\;\;\;\;\dfrac{{\sqrt 5 + 2}}{{\sqrt 5 - 2}} = \dfrac{{{{\left[ {\sqrt 5 + 2} \right]}^2}}}{{5 - {2^2}}} \\\;\;\;= \dfrac{{5 + 2.2\sqrt 5 + {2^2}}}{1} = 9 + 4\sqrt 5 \\\;\;\;\;\dfrac{{\sqrt 7 - \sqrt 5 }}{{\sqrt 7 + \sqrt 5 }} = \dfrac{{{{\left[ {\sqrt 7 - \sqrt 5 } \right]}^2}}}{{7 - 5}}\\\;\;\; = \dfrac{{7 - 2.\sqrt {5.7} + 5}}{2} = 6 - \sqrt {35} .\end{array}\]
\[\begin{array}{l}d]\;\dfrac{{y + a\sqrt y }}{{a\sqrt y }} = \dfrac{{\sqrt y \left[ {\sqrt y + a} \right]}}{{a\sqrt y }} = \dfrac{{\sqrt y + a}}{a}\\\;\;\;\dfrac{{b - \sqrt b }}{{\sqrt b - 1}} = \dfrac{{\sqrt b \left[ {\sqrt b - 1} \right]}}{{\sqrt b - 1}} = \sqrt b .\end{array}\]
\[\begin{array}{l}e]\;\dfrac{b}{{5 + \sqrt b }} = \dfrac{{b\left[ {\sqrt b - 5} \right]}}{{b - {5^2}}} = \dfrac{{b\left[ {\sqrt b - b} \right]}}{{b - 25}}\\\;\;\;\;\dfrac{p}{{2\sqrt p - 1}} = \dfrac{{p\left[ {2\sqrt p + 1} \right]}}{{{{\left[ {2\sqrt p } \right]}^2} - 1}} = \dfrac{{2p\sqrt p + p}}{{4p - 1}}\\\;\;\;\;\dfrac{{a + \sqrt b }}{{\sqrt a - \sqrt b }} = \dfrac{{\left[ {a + \sqrt b } \right]\left[ {\sqrt a + \sqrt b } \right]}}{{a - b}}\end{array}\]