Đề bài
Tính :
a] \[\dfrac{{\sqrt {10} - \sqrt 2 }}{{\sqrt 5 - 1}} + \dfrac{{2 - \sqrt 2 }}{{1 - \sqrt 2 }}\];
b] \[\left[ {\dfrac{{3\sqrt {125} }}{{15}} - \dfrac{{10 - 4\sqrt 5 }}{{\sqrt 5 - 2}}} \right]\dfrac{1}{{\sqrt 5 }}\];
c] \[\dfrac{2}{{\sqrt 7 + \sqrt 3 }} + \sqrt {\dfrac{2}{{5 - \sqrt {21} }}} \];
d] \[\sqrt {\dfrac{{\sqrt 3 }}{{8\sqrt 3 + 3\sqrt {21} }}} \left[ {3\sqrt 2 + \sqrt {14} } \right]\].
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
\[\begin{array}{l}a]\;\dfrac{{\sqrt {10} - \sqrt 2 }}{{\sqrt 5 - 1}} + \dfrac{{2 - \sqrt 2 }}{{1 - \sqrt 2 }}\\ = \dfrac{{\sqrt 2 \left[ {\sqrt 5 - 1} \right]}}{{\sqrt 5 - 1}} + \dfrac{{\sqrt 2 \left[ {\sqrt 2 - 1} \right]}}{{1 - \sqrt 2 }}\\ = \sqrt 2 - \sqrt 2 = 0.\end{array}\]
\[\begin{array}{l}b]\;\left[ {\dfrac{{3\sqrt {125} }}{{15}} - \dfrac{{10 - 4\sqrt 5 }}{{\sqrt 5 - 2}}} \right]\dfrac{1}{{\sqrt 5 }}\\ = \left[ {\dfrac{{3\sqrt {{5^2}.5} }}{{15}} - \dfrac{{2\sqrt 5 \left[ {\sqrt 5 - 2} \right]}}{{\sqrt 5 - 2}}} \right].\dfrac{1}{{\sqrt 5 }}\\ = \left[ {\dfrac{{3.5\sqrt 5 }}{{15}} - 2\sqrt 5 } \right].\dfrac{1}{{\sqrt 5 }}\\ = \left[ {\sqrt 5 - 2\sqrt 5 } \right].\dfrac{1}{{\sqrt 5 }}\\ = - \sqrt 5 .\dfrac{1}{{\sqrt 5 }} = - 1.\end{array}\]
\[\begin{array}{l}c]\;\dfrac{2}{{\sqrt 7 + \sqrt 3 }} + \sqrt {\dfrac{2}{{5 - \sqrt {21} }}} \\ = \dfrac{{2\left[ {\sqrt 7 - \sqrt 3 } \right]}}{{{{\left[ {\sqrt 7 } \right]}^2} - {{\left[ {\sqrt 3 } \right]}^2}}} + \sqrt {\dfrac{{2\left[ {5 + \sqrt {21} } \right]}}{{{5^2} - {{\left[ {\sqrt {21} } \right]}^2}}}} \\ = \dfrac{{2\left[ {\sqrt 7 - \sqrt 3 } \right]}}{{7 - 3}} + \sqrt {\dfrac{{10 + 2\sqrt {21} }}{4}} \\ = \dfrac{{2\left[ {\sqrt 7 - \sqrt 3 } \right]}}{4} + \dfrac{{\sqrt {{{\left[ {\sqrt 7 } \right]}^2} + 2\sqrt 7 .\sqrt 3 + {{\left[ {\sqrt 3 } \right]}^2}} }}{2}\\ = \dfrac{{\sqrt 7 - \sqrt 3 }}{2} + \dfrac{{\sqrt {{{\left[ {\sqrt 7 + \sqrt 3 } \right]}^2}} }}{2}\\ = \dfrac{{\sqrt 7 - \sqrt 3 }}{2} + \dfrac{{\left| {\sqrt 7 + \sqrt 3 } \right|}}{2}\\ = \dfrac{{\sqrt 7 - \sqrt 3 }}{2} + \dfrac{{\sqrt 7 + \sqrt 3 }}{2}\\ = \dfrac{{\sqrt 7 - \sqrt 3 + \sqrt 7 + \sqrt 3 }}{2}\\ = \dfrac{{2\sqrt 7 }}{2} = \sqrt 7 .\end{array}\]
\[\begin{array}{l}d]\;\sqrt {\dfrac{{\sqrt 3 }}{{8\sqrt 3 + 3\sqrt {21} }}} \left[ {3\sqrt 2 + \sqrt {14} } \right]\\ = \sqrt {\dfrac{{\sqrt 3 }}{{\sqrt 3 \left[ {8 + 3\sqrt 7 } \right]}}} .\sqrt 2 \left[ {3 + \sqrt 7 } \right]\\ = \sqrt {\dfrac{1}{{8 + 3\sqrt 7 }}} .\sqrt 2 \left[ {3 + \sqrt 7 } \right]\\ = \sqrt 2 \sqrt {\dfrac{{8 - 3\sqrt 7 }}{{{8^2} - {{\left[ {3\sqrt 7 } \right]}^2}}}} .\left[ {3 + \sqrt 7 } \right]\\ = \sqrt {\dfrac{{16 - 6\sqrt 7 }}{{64 - 63}}} .\left[ {3 + \sqrt 7 } \right]\\ = \sqrt {16 - 6\sqrt 7 } \left[ {3 + \sqrt 7 } \right]\\ = \sqrt {{3^2} - 2.3\sqrt 7 + {{\left[ {\sqrt 7 } \right]}^2}} \left[ {3 + \sqrt 7 } \right]\\ = \sqrt {{{\left[ {3 - \sqrt 7 } \right]}^2}} \left[ {3 + \sqrt 7 } \right]\\ = \left| {3 - \sqrt 7 } \right|\left[ {3 + \sqrt 7 } \right]\\ = \left[ {3 - \sqrt 7 } \right]\left[ {3 + \sqrt 7 } \right]\\ = {3^2} - 7 = 9 - 7 = 2.\end{array}\]