Đáp án:
n. 3
o. 3
Giải thích các bước giải:
\(\begin{array}{l}
B6:\\
c.\dfrac{{\sqrt 2 .\sqrt 5 + \sqrt 3 .\sqrt 5 }}{{2\sqrt 2 + 2\sqrt 3 }} = \dfrac{{\sqrt 5 \left( {\sqrt 2 + \sqrt 3 } \right)}}{{2\left( {\sqrt 2 + \sqrt 3 } \right)}}\\
= \dfrac{{\sqrt 5 }}{2}\\
d.\dfrac{{\sqrt 2 \left( {\sqrt 2 + \sqrt 3 } \right)}}{{\sqrt 2 + \sqrt 3 }} = \sqrt 2 \\
j.\sqrt {6 - 2.\sqrt 6 .1 + 1} = \sqrt {{{\left( {\sqrt 6 - 1} \right)}^2}} \\
= \sqrt 6 - 1\\
k.\dfrac{{\sqrt {3 - 2\sqrt 3 .1 + 1} }}{{\sqrt 2 \left( {\sqrt 3 - 1} \right)}} = \dfrac{{\sqrt {{{\left( {\sqrt 3 - 1} \right)}^2}} }}{{\sqrt 2 \left( {\sqrt 3 - 1} \right)}}\\
= \dfrac{{\sqrt 3 - 1}}{{\sqrt 2 \left( {\sqrt 3 - 1} \right)}}\\
= \dfrac{1}{{\sqrt 2 }}\\
l.\left( {2 + \sqrt 3 } \right).\sqrt {4 - 2.2.\sqrt 3 + 3} \\
= \left( {2 + \sqrt 3 } \right)\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} \\
= \left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)\\
= 4 - 3 = 1\\
m.\sqrt {3 + 2\sqrt 3 .1 + 1} - \sqrt {3 + 2\sqrt 3 .\sqrt 2 + 2} \\
= \sqrt {{{\left( {\sqrt 3 + 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}} \\
= \sqrt 3 + 1 - \sqrt 3 - \sqrt 2 \\
= 1 - \sqrt 2 \\
n.\left( {\sqrt {3 - 2\sqrt 3 .\sqrt 2 + 2} + \sqrt 2 } \right).\sqrt 3 \\
= \left( {\sqrt {{{\left( {\sqrt 3 - \sqrt 2 } \right)}^2}} + \sqrt 2 } \right).\sqrt 3 \\
= \left( {\sqrt 3 - \sqrt 2 + \sqrt 2 } \right).\sqrt 3 = 3\\
o.\sqrt {2 + 2\sqrt 2 .1 + 1} + \sqrt {4 - 2.2.\sqrt 2 + 2} \\
= \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} + \sqrt {{{\left( {2 - \sqrt 2 } \right)}^2}} \\
= \sqrt 2 + 1 + 2 - \sqrt 2 \\
= 3
\end{array}\)
