Đáp án:
\(\begin{array}{l}
B = 2\sqrt 2 \\
C = 4\sqrt 5 \\
D = 2\sqrt 5 \\
E = 1\\
F = b - a
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
B = \sqrt {2 - 2\sqrt 2 .1 + 1} + \sqrt {2 + 2\sqrt 2 .1 + 1} \\
= \sqrt {{{\left( {\sqrt 2 - 1} \right)}^2}} + \sqrt {{{\left( {\sqrt 2 + 1} \right)}^2}} \\
= \sqrt 2 - 1 + \sqrt 2 + 1 = 2\sqrt 2 \\
C = 2\sqrt 5 - 5\sqrt 5 - 4\sqrt 5 + 11\sqrt 5 \\
= \left( {2 - 5 - 4 + 11} \right)\sqrt 5 \\
= 4\sqrt 5 \\
D = \dfrac{{2\sqrt 5 \left( {\sqrt 5 + \sqrt 2 } \right)}}{{\sqrt 5 + \sqrt 2 }} = 2\sqrt 5 \\
E = \left[ {2 + \dfrac{{\sqrt 3 \left( {\sqrt 3 + 1} \right)}}{{\sqrt 3 + 1}}} \right].\left[ {2 - \dfrac{{\sqrt 3 \left( {\sqrt 3 - 1} \right)}}{{\sqrt 3 - 1}}} \right]\\
= \left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right)\\
= 4 - 3 = 1\\
F = \left[ {\dfrac{{\sqrt b }}{{\sqrt a \left( {\sqrt a - \sqrt b } \right)}} - \dfrac{{\sqrt a }}{{\sqrt b \left( {\sqrt a - \sqrt b } \right)}}} \right].\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)\\
= \dfrac{{b - a}}{{\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)}}.\sqrt {ab} \left( {\sqrt a - \sqrt b } \right)\\
= b - a
\end{array}\)
