Đáp án:
\(\begin{array}{l}
A = 1\\
B = 3\\
C = 5\\
D = \sqrt {14 + 2\sqrt 5 }
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
A = \sqrt 5 - \sqrt {3 - \sqrt {20 - 2.2\sqrt 5 .3 + 9} } \\
= \sqrt 5 - \sqrt {3 - \sqrt {{{\left( {2\sqrt 5 - 3} \right)}^2}} } \\
= \sqrt 5 - \sqrt {3 - \left( {2\sqrt 5 - 3} \right)} \\
= \sqrt 5 - \sqrt {3 - 2\sqrt 5 + 3} \\
= \sqrt 5 - \sqrt {5 - 2\sqrt 5 .1 + 1} \\
= \sqrt 5 - \sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
= \sqrt 5 - \sqrt 5 + 1 = 1\\
B = \dfrac{{\sqrt 3 + \sqrt {9 + 2.3.\sqrt 2 + 2} - \sqrt {3 + 2\sqrt 3 .\sqrt 2 + 2} }}{{\sqrt 2 + \sqrt {5 + 2\sqrt 5 .1 + 1} - \sqrt {5 + 2\sqrt 5 .\sqrt 2 + 2} }}\\
= \dfrac{{\sqrt 3 + \sqrt {{{\left( {3 + \sqrt 2 } \right)}^2}} - \sqrt {{{\left( {\sqrt 3 + \sqrt 2 } \right)}^2}} }}{{\sqrt 2 + \sqrt {{{\left( {\sqrt 5 + 1} \right)}^2}} - \sqrt {{{\left( {\sqrt 5 + \sqrt 2 } \right)}^2}} }}\\
= \dfrac{{\sqrt 3 + 3 + \sqrt 2 - \sqrt 3 - \sqrt 2 }}{{\sqrt 2 + \sqrt 5 + 1 - \sqrt 5 - \sqrt 2 }}\\
= \dfrac{3}{1} = 3\\
C = \sqrt {5\sqrt 3 + 5\sqrt {48 - 10\sqrt {4 + 2.2.\sqrt 3 + 3} } } \\
= \sqrt {5\sqrt 3 + 5\sqrt {48 - 10\sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} } } \\
= \sqrt {5\sqrt 3 + 5\sqrt {48 - 10\left( {2 + \sqrt 3 } \right)} } \\
= \sqrt {5\sqrt 3 + 5\sqrt {48 - 20 - 10\sqrt 3 } } \\
= \sqrt {5\sqrt 3 + 5\sqrt {28 - 10\sqrt 3 } } \\
= \sqrt {5\sqrt 3 + 5\sqrt {25 - 2.5.\sqrt 3 + 3} } \\
= \sqrt {5\sqrt 3 + 5\sqrt {{{\left( {5 - \sqrt 3 } \right)}^2}} } \\
= \sqrt {5\sqrt 3 + 5\left( {5 - \sqrt 3 } \right)} \\
= \sqrt {5\sqrt 3 + 25 - 5\sqrt 3 } = \sqrt {25} = 5\\
D = \sqrt {4 + \sqrt {10 + 2\sqrt 5 } } + \sqrt {4 - \sqrt {10 + 2\sqrt 5 } } \\
\to {D^2} = 4 + \sqrt {10 + 2\sqrt 5 } + 2\sqrt {\left( {4 + \sqrt {10 + 2\sqrt 5 } } \right)\left( {4 - \sqrt {10 + 2\sqrt 5 } } \right)} + 4 - \sqrt {10 + 2\sqrt 5 } \\
= 16 + 2\sqrt {16 - \left( {10 + 2\sqrt 5 } \right)} \\
= 16 + 2\sqrt {6 - 2\sqrt 5 } \\
= 16 + 2\sqrt {5 - 2.\sqrt 5 .1 + 1} \\
= 16 + 2\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
= 16 + 2\left( {\sqrt 5 - 1} \right)\\
= 14 + 2\sqrt 5 \\
\to D = \sqrt {14 + 2\sqrt 5 }
\end{array}\)
