Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
\left( {\sqrt {28}  - 2\sqrt {14}  + \sqrt 7 } \right).\sqrt 7  + 7\sqrt 8 \\
 = \left( {\sqrt {{2^2}.7}  - 2.\sqrt {2.7}  + \sqrt 7 } \right).\sqrt 7  + 7\sqrt 8 \\
 = \left( {2\sqrt 7  - 2.\sqrt {2.7}  + \sqrt 7 } \right).\sqrt 7  + 7\sqrt {{2^2}.2} \\
 = \left( {3\sqrt 7  - 2\sqrt {2.7} } \right).\sqrt 7  + 7.2\sqrt 2 \\
 = 3{\sqrt 7 ^2} - 2.\sqrt 2 .{\sqrt 7 ^2} + 14\sqrt 2 \\
 = 21 - 14\sqrt 2  + 14\sqrt 2 \\
 = 21\\
2,\\
\left( {\dfrac{1}{{2\sqrt 5  - \sqrt 3 }} + \dfrac{1}{{2\sqrt 5  + \sqrt 3 }}} \right):\dfrac{{\sqrt 3  + 1}}{{17}}\\
 = \dfrac{{\left( {2\sqrt 5  + \sqrt 3 } \right) + \left( {2\sqrt 5  - \sqrt 3 } \right)}}{{\left( {2\sqrt 5  - \sqrt 3 } \right)\left( {2\sqrt 5  + \sqrt 3 } \right)}}:\dfrac{{\sqrt 3  + 1}}{{17}}\\
 = \dfrac{{4\sqrt 5 }}{{{{\left( {2\sqrt 5 } \right)}^2} - {{\left( {\sqrt 3 } \right)}^2}}}:\dfrac{{\sqrt 3  + 1}}{{17}}\\
 = \dfrac{{4\sqrt 5 }}{{20 - 3}}.\dfrac{{17}}{{\sqrt 3  + 1}}\\
 = \dfrac{{4\sqrt 5 }}{{17}}.\dfrac{{17}}{{\sqrt 3  + 1}}\\
 = \dfrac{{4\sqrt 5 }}{{\sqrt 3  + 1}}\\
3,\\
a,\\
A = \left( {1 + \dfrac{{\sqrt a }}{{a + 1}}} \right):\left( {\dfrac{1}{{\sqrt a  - 1}} - \dfrac{{2\sqrt a }}{{a\sqrt a  + \sqrt a  - a - 1}}} \right)\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}
a \ge 0\\
a \ne 1
\end{array} \right)\\
 = \dfrac{{a + \sqrt a  + 1}}{{a + 1}}:\left( {\dfrac{1}{{\sqrt a  - 1}} - \dfrac{{2\sqrt a }}{{\sqrt a \left( {a + 1} \right) - \left( {a + 1} \right)}}} \right)\\
 = \dfrac{{a + \sqrt a  + 1}}{{a + 1}}:\left( {\dfrac{1}{{\sqrt a  - 1}} - \dfrac{{2\sqrt a }}{{\left( {a + 1} \right)\left( {\sqrt a  - 1} \right)}}} \right)\\
 = \dfrac{{a + \sqrt a  + 1}}{{a + 1}}:\dfrac{{\left( {a + 1} \right) - 2\sqrt a }}{{\left( {\sqrt a  - 1} \right)\left( {a + 1} \right)}}\\
 = \dfrac{{a + \sqrt a  + 1}}{{a + 1}}:\dfrac{{{{\left( {\sqrt a  - 1} \right)}^2}}}{{\left( {\sqrt a  - 1} \right).\left( {a + 1} \right)}}\\
 = \dfrac{{a + \sqrt a  + 1}}{{a + 1}}.\dfrac{{\left( {\sqrt a  - 1} \right)\left( {a + 1} \right)}}{{{{\left( {\sqrt a  - 1} \right)}^2}}}\\
 = \dfrac{{a + \sqrt a  + 1}}{{\sqrt a  - 1}}\\
b,\\
a = 4 + 2\sqrt 3  = 3 + 2\sqrt 3 .1 + 1 = {\left( {\sqrt 3  + 1} \right)^2}\\
 \Rightarrow \sqrt a  = \sqrt 3  + 1\\
A = \dfrac{{\left( {4 + 2\sqrt 3 } \right) + \left( {\sqrt 3  + 1} \right) + 1}}{{\left( {\sqrt 3  + 1} \right) - 1}} = \dfrac{{6 + 3\sqrt 3 }}{{\sqrt 3 }} = 3 + 2\sqrt 3 \\
c,\\
A > 1 \Leftrightarrow \dfrac{{a + \sqrt a  + 1}}{{\sqrt a  - 1}} > 1\\
 \Leftrightarrow \dfrac{{a + \sqrt a  + 1}}{{\sqrt a  - 1}} - 1 > 0\\
 \Leftrightarrow \dfrac{{\left( {a + \sqrt a  + 1} \right) - \left( {\sqrt a  - 1} \right)}}{{\sqrt a  - 1}} > 0\\
 \Leftrightarrow \dfrac{{a + 2}}{{\sqrt a  - 1}} > 0\\
 \Leftrightarrow \sqrt a  - 1 > 0\\
 \Leftrightarrow \sqrt a  > 1\\
 \Leftrightarrow a > 1
\end{array}\)