Đáp án:
\(\begin{array}{l}
27)\sqrt 6 \\
29)\sqrt 6 \\
28)1\\
30)\dfrac{{32}}{9}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
27)\dfrac{1}{{\sqrt {2 - \sqrt 3 } }} + \dfrac{1}{{\sqrt {2 + \sqrt 3 } }}\\
= \dfrac{{\sqrt {2 + \sqrt 3 } + \sqrt {2 - \sqrt 3 } }}{{\sqrt {\left( {2 - \sqrt 3 } \right)\left( {2 + \sqrt 3 } \right)} }}\\
= \dfrac{{\sqrt {2 + \sqrt 3 } + \sqrt {2 - \sqrt 3 } }}{{\sqrt {4 - 3} }}\\
= \sqrt {2 + \sqrt 3 } + \sqrt {2 - \sqrt 3 } \\
= \dfrac{{\sqrt {4 + 2\sqrt 3 } + \sqrt {4 - 2\sqrt 3 } }}{{\sqrt 2 }}\\
= \dfrac{{\sqrt {3 + 2\sqrt 3 .1 + 1} + \sqrt {3 - 2\sqrt 3 .1 + 1} }}{{\sqrt 2 }}\\
= \dfrac{{\sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} }}{{\sqrt 2 }}\\
= \dfrac{{\sqrt 3 + 1 + \sqrt 3 - 1}}{{\sqrt 2 }}\\
= \dfrac{{2\sqrt 3 }}{{\sqrt 2 }} = \sqrt 6 \\
29)\sqrt {\dfrac{{\left( {3\sqrt 3 - 4} \right)\left( {2\sqrt 3 - 1} \right)}}{{\left( {2\sqrt 3 - 1} \right)\left( {2\sqrt 3 + 1} \right)}}} + \sqrt {\dfrac{{\left( {\sqrt 3 + 4} \right)\left( {5 + 2\sqrt 3 } \right)}}{{\left( {5 + 2\sqrt 3 } \right)\left( {5 - 2\sqrt 3 } \right)}}} \\
= \sqrt {\dfrac{{6.3 - 3\sqrt 3 - 8\sqrt 3 + 4}}{{12 - 1}}} + \sqrt {\dfrac{{5\sqrt 3 + 2.3 + 20 + 8\sqrt 3 }}{{25 - 12}}} \\
= \sqrt {\dfrac{{22 - 11\sqrt 3 }}{{11}}} + \sqrt {\dfrac{{26 + 13\sqrt 3 }}{{13}}} \\
= \sqrt {2 - \sqrt 3 } + \sqrt {2 + \sqrt 3 } \\
= \dfrac{{\sqrt {4 - 2\sqrt 3 } + \sqrt {4 + 2\sqrt 3 } }}{{\sqrt 2 }}\\
= \dfrac{{\sqrt {3 - 2\sqrt 3 .1 + 1} + \sqrt {3 + 2\sqrt 3 .1 + 1} }}{{\sqrt 2 }}\\
= \dfrac{{\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} }}{{\sqrt 2 }}\\
= \dfrac{{\sqrt 3 - 1 + \sqrt 3 + 1}}{{\sqrt 2 }}\\
= \dfrac{{2\sqrt 3 }}{{\sqrt 2 }} = \sqrt 6 \\
28)\dfrac{{\sqrt {3 - \sqrt 5 } \left( {3 + \sqrt 5 } \right)}}{{\sqrt 2 \left( {\sqrt 5 + 1} \right)}}\\
= \dfrac{{\sqrt {6 - 2\sqrt 5 } \left( {3 + \sqrt 5 } \right)}}{{2\left( {\sqrt 5 + 1} \right)}}\\
= \dfrac{{\sqrt {5 - 2.\sqrt 5 .1 + 1} .\left( {3 + \sqrt 5 } \right)}}{{2\left( {\sqrt 5 + 1} \right)}}\\
= \dfrac{{\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} .\left( {3 + \sqrt 5 } \right)}}{{2\left( {\sqrt 5 + 1} \right)}}\\
= \dfrac{{\left( {\sqrt 5 - 1} \right)\left( {3 + \sqrt 5 } \right)}}{{2\left( {\sqrt 5 + 1} \right)}}\\
= \dfrac{{3\sqrt 5 + 5 - 3 - \sqrt 5 }}{{2\left( {\sqrt 5 + 1} \right)}}\\
= \dfrac{{2\sqrt 5 + 2}}{{2\left( {\sqrt 5 + 1} \right)}}\\
= \dfrac{{\sqrt 5 + 1}}{{\sqrt 5 + 1}} = 1\\
30)\dfrac{{5 + \sqrt 7 }}{{9 - \sqrt {16 + 2.4.\sqrt 7 + 7} }} + \dfrac{{5 - \sqrt 7 }}{{2 + \sqrt {9 + 2.3.\sqrt 7 + 7} }}\\
= \dfrac{{5 + \sqrt 7 }}{{9 - \sqrt {{{\left( {4 + \sqrt 7 } \right)}^2}} }} + \dfrac{{5 - \sqrt 7 }}{{2 + \sqrt {{{\left( {3 + \sqrt 7 } \right)}^2}} }}\\
= \dfrac{{5 + \sqrt 7 }}{{9 - \left( {4 + \sqrt 7 } \right)}} + \dfrac{{5 - \sqrt 7 }}{{2 + \left( {3 + \sqrt 7 } \right)}}\\
= \dfrac{{5 + \sqrt 7 }}{{5 - \sqrt 7 }} + \dfrac{{5 - \sqrt 7 }}{{5 + \sqrt 7 }}\\
= \dfrac{{{{\left( {5 + \sqrt 7 } \right)}^2} + {{\left( {5 - \sqrt 7 } \right)}^2}}}{{25 - 7}}\\
= \dfrac{{25 + 10\sqrt 7 + 7 + 25 - 10\sqrt 7 + 7}}{{18}}\\
= \dfrac{{64}}{{18}} = \dfrac{{32}}{9}
\end{array}\)
( câu 30 t nghĩ đổi đề thành \({2 + \sqrt {16 + 6\sqrt 7 } }\) thì hợp lý hơn nha )
