Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
A = \sqrt {8 + 2\sqrt {10 + 2\sqrt 5 } } + \sqrt {8 - 2\sqrt {10 + 2\sqrt 5 } } \\
\Leftrightarrow {A^2} = 8 + 2\sqrt {10 + 2\sqrt 5 } + 2\sqrt {\left( {8 + 2\sqrt {10 + 2\sqrt 5 } } \right)\left( {8 - 2\sqrt {10 + 2\sqrt 5 } } \right)} + 8 - 2\sqrt {10 + 2\sqrt 5 } \\
\Leftrightarrow {A^2} = 16 + 2\sqrt {\left( {8 + 2\sqrt {10 + 2\sqrt 5 } } \right)\left( {8 - 2\sqrt {10 + 2\sqrt 5 } } \right)} \\
\Leftrightarrow {A^2} = 16 + 2\sqrt {64 - 4.\left( {10 + 2\sqrt 5 } \right)} \\
\Leftrightarrow {A^2} = 16 + 2\sqrt {24 - 8\sqrt 5 } \\
\Leftrightarrow {A^2} = 16 + 2\sqrt {4.\left( {6 - 2\sqrt 5 } \right)} \\
\Leftrightarrow {A^2} = 16 + 4\sqrt {6 - 2\sqrt 5 } \\
\Leftrightarrow {A^2} = 16 + 4\sqrt {{{\left( {\sqrt 5 - 1} \right)}^2}} \\
\Leftrightarrow {A^2} = 16 + 4.\left( {\sqrt 5 - 1} \right)\\
\Leftrightarrow {A^2} = 12 + 4\sqrt 5 \\
\Leftrightarrow {A^2} = 2.\left( {6 + 2\sqrt 5 } \right)\\
\Leftrightarrow {A^2} = 2.{\left( {\sqrt 5 + 1} \right)^2}\\
A > 0 \Rightarrow A = \sqrt 2 .\left( {\sqrt 5 + 1} \right) = \sqrt {10} + \sqrt 2 \\
b,\\
B = \left( {\frac{{a\sqrt a - 3}}{{a - 2\sqrt a - 3}} - \frac{{2\left( {\sqrt a - 3} \right)}}{{\sqrt a + 1}}} \right):\left( {\frac{{a + 8}}{{a - 1}}} \right)\\
= \left( {\frac{{a\sqrt a - 3}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a + 1} \right)}} - \frac{{2{{\left( {\sqrt a - 3} \right)}^2}}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a + 1} \right)}}} \right):\frac{{a + 8}}{{a - 1}}\\
= \left( {\frac{{a\sqrt a - 3 - 2\left( {a - 6\sqrt a + 9} \right)}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a + 1} \right)}}} \right):\frac{{a + 8}}{{a - 1}}\\
= \frac{{a\sqrt a - 2a + 12\sqrt a - 21}}{{\left( {\sqrt a - 3} \right)\left( {\sqrt a + 1} \right)}}:\frac{{a + 8}}{{a - 1}}\\
= ???
\end{array}\)
(Em xem lại đề câu b nhé!)
