Đáp án:
\(\begin{array}{l}
a)6\\
b) - \sqrt {2013} + \sqrt {2015} \\
c)2\sqrt {10} \\
d)\sqrt 5 - 1
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
a)\left( {3\sqrt 2 + 2\sqrt 3 } \right)\left( {2\sqrt 3 - 3\sqrt 2 } \right)\\
= {\left( {3\sqrt 2 } \right)^2} - {\left( {2\sqrt 3 } \right)^2}\\
= 18 - 12 = 6\\
b)\dfrac{{\sqrt {2013} + \sqrt {2014} }}{{2013 - 2014}} - \dfrac{{\sqrt {2014} + \sqrt {2015} }}{{2014 - 2015}}\\
= - \sqrt {2013} - \sqrt {2014} + \sqrt {2014} + \sqrt {2015} \\
= - \sqrt {2013} + \sqrt {2015} \\
c)\left| {4 + \sqrt {10} } \right| - \left| {4 - \sqrt {10} } \right|\\
= 4 + \sqrt {10} - \left( {4 - \sqrt {10} } \right)\left( {DK:4 > \sqrt {10} } \right)\\
= 4 + \sqrt {10} - 4 + \sqrt {10} = 2\sqrt {10} \\
d)\sqrt {2 - 2\sqrt 2 .1 + 1} + \sqrt {4 - 2.2\sqrt 2 + 2} + \sqrt {5 - 2.2\sqrt 5 + 4} \\
= \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} + \sqrt {{{\left( {2 - \sqrt 2 } \right)}^2}} + \sqrt {{{\left( {\sqrt 5 - 2} \right)}^2}} \\
= \sqrt 2 - 1 + 2 - \sqrt 2 + \sqrt 5 - 2\\
= \sqrt 5 - 1
\end{array}\)
