Đáp án:
\(dpcm\)
Giải thích các bước giải:
\(\begin{array}{l}
c)VT = 2\sqrt 6 - 4\sqrt 2 + 1 + 4\sqrt 2 + 8 - 2\sqrt 6 \\
= 1 + 8 = 9 = VP\\
e)VT = \left( {3 + \sqrt 5 } \right).\sqrt 2 \left( {\sqrt 5 - 1} \right).\sqrt {3 - \sqrt 5 } \\
= \left( {3 + \sqrt 5 } \right)\left( {\sqrt 5 - 1} \right)\sqrt {6 - 2\sqrt 5 } \\
= \left( {3 + \sqrt 5 } \right)\left( {\sqrt 5 - 1} \right)\sqrt {5 - 2\sqrt 5 .1 + 1} \\
= \left( {3 + \sqrt 5 } \right)\left( {\sqrt 5 - 1} \right)\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
= \left( {3 + \sqrt 5 } \right){\left( {\sqrt 5 - 1} \right)^2}\\
= \left( {3 + \sqrt 5 } \right)\left( {6 - 2\sqrt 5 } \right)\\
= 2\left( {3 + \sqrt 5 } \right)\left( {3 - \sqrt 5 } \right)\\
= 2\left( {9 - 5} \right) = 2.4 = 8 = VP\\
d)VT = \dfrac{2}{{\left| {2 - \sqrt 5 } \right|}} - \dfrac{2}{{\left| {2 + \sqrt 5 } \right|}}\\
= \dfrac{2}{{\sqrt 5 - 2}} - \dfrac{2}{{\sqrt 5 + 2}}\\
= \dfrac{{2\sqrt 5 + 4 - 2\sqrt 5 + 4}}{{5 - 4}} = 8 = VP\\
f)\sqrt {\sqrt 2 + 1} - \sqrt {\sqrt 2 - 1} = \sqrt {2\left( {\sqrt 2 - 1} \right)} \\
\Leftrightarrow \sqrt 2 + 1 - 2\sqrt {\left( {\sqrt 2 + 1} \right)\left( {\sqrt 2 - 1} \right)} + \sqrt 2 - 1 = 2\left( {\sqrt 2 - 1} \right)\\
\Leftrightarrow 2\sqrt 2 - 2\sqrt {2 - 1} = 2\sqrt 2 - 2\\
\Leftrightarrow 2\sqrt 2 - 2 = 2\sqrt 2 - 2\left( {ld} \right)\\
\to dpcm
\end{array}\)