Đáp án:
\(c.\sqrt {\sqrt 2 + 1} - \sqrt {\sqrt 2 - 1} = \sqrt {2\left( {\sqrt 2 - 1} \right)} \)
Giải thích các bước giải:
\(\begin{array}{l}
a.VT = \dfrac{{\sqrt {xy} \left( {\sqrt x + \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)}}{{\sqrt {xy} }}\\
= \left( {\sqrt x + \sqrt y } \right)\left( {\sqrt x + \sqrt y } \right)\\
= x - y = VP\\
b.VT = \left( {3 + \sqrt 5 } \right).\sqrt 2 \left( {\sqrt 5 - 1} \right)\sqrt {3 - \sqrt 5 } \\
= \left( {3 + \sqrt 5 } \right)\left( {\sqrt 5 - 1} \right)\sqrt {6 - 2\sqrt 5 } \\
= \left( {3 + \sqrt 5 } \right)\left( {\sqrt 5 - 1} \right)\sqrt {5 - 2.\sqrt 5 .1 + 1} \\
= \left( {3 + \sqrt 5 } \right)\left( {\sqrt 5 - 1} \right)\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
= \left( {3 + \sqrt 5 } \right){\left( {\sqrt 5 - 1} \right)^2}\\
= \left( {3 + \sqrt 5 } \right)\left( {5 - 2\sqrt 5 + 1} \right)\\
= \left( {3 + \sqrt 5 } \right)\left( {6 - 2\sqrt 5 } \right)\\
= 2\left( {3 + \sqrt 5 } \right)\left( {3 - \sqrt 5 } \right)\\
= 2.\left( {9 - 5} \right) = 2.4 = 8 = VP\\
c.\sqrt {\sqrt 2 + 1} - \sqrt {\sqrt 2 - 1} = \sqrt {2\left( {\sqrt 2 - 1} \right)} \\
\to \sqrt 2 + 1 - 2\sqrt {\sqrt 2 + 1} .\sqrt {\sqrt 2 - 1} + \sqrt 2 - 1 = 2\left( {\sqrt 2 - 1} \right)\\
\to 2\sqrt 2 - 2\sqrt {2 - 1} = 2\sqrt 2 - 2\\
\to 2\sqrt 2 - 2 = 2\sqrt 2 - 2\left( {ld} \right)\\
\to dpcm
\end{array}\)
