Đáp án:
\(\begin{array}{l}
B3:\\
a)\dfrac{{5\sqrt 5 - 5\sqrt 2 + 10 - 2\sqrt {10} }}{3}\\
c)\dfrac{{2\sqrt 2 + 3\sqrt 3 }}{{ - 19}}\\
b)\dfrac{{\sqrt 2 }}{2}\\
d)\dfrac{{3 - \sqrt 5 }}{2}\\
B4:\\
a)\sqrt 2 \\
c) - 1\\
b)4\\
d)\sqrt 5 - \sqrt 2 + \sqrt 3 - 2
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B3:\\
a)\dfrac{{\left( {5 + 2\sqrt 5 } \right)\left( {\sqrt 5 - \sqrt 2 } \right)}}{{\left( {\sqrt 5 - \sqrt 2 } \right)\left( {\sqrt 5 + \sqrt 2 } \right)}}\\
= \dfrac{{5\sqrt 5 - 5\sqrt 2 + 10 - 2\sqrt {10} }}{{5 - 2}}\\
= \dfrac{{5\sqrt 5 - 5\sqrt 2 + 10 - 2\sqrt {10} }}{3}\\
c)\dfrac{{2\sqrt 2 + 3\sqrt 3 }}{{8 - 27}} = \dfrac{{2\sqrt 2 + 3\sqrt 3 }}{{ - 19}}\\
b)\dfrac{{\sqrt 2 \left( {\sqrt {12} - \sqrt 5 } \right)}}{{2\left( {\sqrt {12} - \sqrt 5 } \right)}} = \dfrac{{\sqrt 2 }}{2}\\
d)\sqrt {\dfrac{{3 - \sqrt 5 }}{{3 + \sqrt 5 }}} = \sqrt {\dfrac{{{{\left( {3 - \sqrt 5 } \right)}^2}}}{{\left( {3 - \sqrt 5 } \right)\left( {3 + \sqrt 5 } \right)}}} \\
= \dfrac{{\left| {3 - \sqrt 5 } \right|}}{{\sqrt {9 - 5} }} = \dfrac{{3 - \sqrt 5 }}{2}\\
B4:\\
a)\dfrac{{\sqrt 3 \left( {\sqrt 3 + 2} \right)}}{{\sqrt 3 }} + \dfrac{{\sqrt 2 \left( {\sqrt 2 + 1} \right)}}{{\sqrt 2 + 1}} - 2 - \sqrt 3 \\
= \sqrt 3 + 2 + \sqrt 2 - 2 - \sqrt 3 \\
= \sqrt 2 \\
c)\left[ {\dfrac{{\sqrt 5 \left( {\sqrt 5 - 2} \right)}}{{ - \left( {\sqrt 5 - 2} \right)}} - 2} \right].\left[ {\dfrac{{\sqrt 5 \left( {\sqrt 5 + 3} \right)}}{{\sqrt 5 + 3}} - 2} \right]\\
= \left( { - \sqrt 5 - 2} \right)\left( {\sqrt 5 - 2} \right)\\
= - \left( {\sqrt 5 + 2} \right)\left( {\sqrt 5 - 2} \right)\\
= - \left( {5 - 4} \right) = - 1\\
b)\left[ {1 - \dfrac{{\sqrt 5 \left( {\sqrt 5 + 1} \right)}}{{\sqrt 5 + 1}}} \right].\left[ {\dfrac{{\sqrt 5 \left( {\sqrt 5 - 1} \right)}}{{ - \left( {\sqrt 5 - 1} \right)}} - 1} \right]\\
= \left( {1 - \sqrt 5 } \right)\left( { - \sqrt 5 - 1} \right)\\
= - \left( {1 - \sqrt 5 } \right)\left( {1 + \sqrt 5 } \right)\\
= - \left( {1 - 5} \right) = 4\\
d)\dfrac{{3\left( {\sqrt 5 + \sqrt 2 } \right)}}{{5 - 2}} - \dfrac{{2\left( {2 + \sqrt 2 } \right)}}{{4 - 2}} + \dfrac{{\sqrt 3 - \sqrt 2 }}{{3 - 2}}\\
= \sqrt 5 + \sqrt 2 - 2 - \sqrt 2 + \sqrt 3 - \sqrt 2 \\
= \sqrt 5 - \sqrt 2 + \sqrt 3 - 2
\end{array}\)
