Đáp án:

\(\begin{array}{l}
1,\\
a,\,\,\, - 2\sqrt 3 \\
b,\,\,\,\,2\\
c,\,\,\,\sqrt 3 \\
2,\\
a,\\
A = \sqrt a  - 1\\
b,\\
A = 1\\
c,\\
0 < a < 1\\
3,\\
a,\\
\left[ \begin{array}{l}
x = 7\\
x =  - 1
\end{array} \right.\\
b,\\
x =  \pm \sqrt 6 \\
c,\\
x = 3
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
a,\\
\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}}  - \sqrt {{{\left( {2 + \sqrt 3 } \right)}^2}} \\
 = \left| {2 - \sqrt 3 } \right| - \left| {2 + \sqrt 3 } \right|\\
 = \left( {2 - \sqrt 3 } \right) - \left( {2 + \sqrt 3 } \right)\\
 = 2 - \sqrt 3  - 2 - \sqrt 3 \\
 =  - 2\sqrt 3 \\
b,\\
\left( {\sqrt {27}  - 3\sqrt {12}  + 5\sqrt 3 } \right):\sqrt 3 \\
 = \left( {\sqrt {9.3}  - 3\sqrt {4.3}  + 5\sqrt 3 } \right):\sqrt 3 \\
 = \left( {\sqrt {{3^2}.3}  - 3.\sqrt {{2^2}.3}  + 5\sqrt 3 } \right):\sqrt 3 \\
 = \left( {3\sqrt 3  - 3.2\sqrt 3  + 5\sqrt 3 } \right):\sqrt 3 \\
 = \left( {3\sqrt 3  - 6\sqrt 3  + 5\sqrt 3 } \right):\sqrt 3 \\
 = 2\sqrt 3 :\sqrt 3 \\
 = 2\\
c,\\
\tan 60^\circ .{\cos ^2}47^\circ  + {\sin ^2}47^\circ .\cot 30^\circ \\
 = \tan 60^\circ .{\cos ^2}47^\circ  + {\sin ^2}47^\circ .\tan \left( {90^\circ  - 30^\circ } \right)\\
 = \tan 60^\circ .{\cos ^2}47^\circ  + {\sin ^2}47^\circ .\tan 60^\circ \\
 = \tan 60^\circ .\left( {{{\cos }^2}47^\circ  + {{\sin }^2}47^\circ } \right)\\
 = \tan 60^\circ .1\\
 = \tan 60^\circ \\
 = \dfrac{{\sin 60^\circ }}{{\cos 60^\circ }}\\
 = \dfrac{{\dfrac{{\sqrt 3 }}{2}}}{{\dfrac{1}{2}}}\\
 = \sqrt 3 \\
2,\\
a,\\
A = \left( {\dfrac{{\sqrt a }}{{\sqrt a  - 1}} - \dfrac{{\sqrt a }}{{a - \sqrt a }}} \right):\dfrac{{\sqrt a  + 1}}{{a - 1}}\\
 = \left( {\dfrac{{\sqrt a }}{{\sqrt a  - 1}} - \dfrac{{\sqrt a }}{{\sqrt a \left( {\sqrt a  - 1} \right)}}} \right):\dfrac{{\sqrt a  + 1}}{{{{\sqrt a }^2} - {1^2}}}\\
 = \left( {\dfrac{{\sqrt a }}{{\sqrt a  - 1}} - \dfrac{1}{{\sqrt a  - 1}}} \right):\dfrac{{\sqrt a  + 1}}{{\left( {\sqrt a  - 1} \right)\left( {\sqrt a  + 1} \right)}}\\
 = \dfrac{{\sqrt a  - 1}}{{\sqrt a  - 1}}:\dfrac{1}{{\sqrt a  - 1}}\\
 = 1:\dfrac{1}{{\sqrt a  - 1}}\\
 = \sqrt a  - 1\\
b,\\
{a^2} - 4a = 0\\
 \Leftrightarrow a.\left( {a - 4} \right) = 0\\
 \Leftrightarrow \left[ \begin{array}{l}
a = 0\\
a - 4 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
a = 0\\
a = 4
\end{array} \right.\\
a > 0,a \ne 0 \Rightarrow a = 4\\
a = 4 \Rightarrow A = \sqrt 4  - 1 = 2 - 1 = 1\\
c,\\
A < 0\\
 \Leftrightarrow \sqrt a  - 1 < 0\\
 \Leftrightarrow \sqrt a  < 1\\
 \Leftrightarrow a < 1\\
a > 0,a \ne 1 \Rightarrow 0 < a < 1\\
3,\\
a,\\
DKXD:\,\,\,{\left( {x - 3} \right)^2} \ge 0,\,\,\,\forall x\\
\sqrt {{{\left( {x - 3} \right)}^2}}  = 4\\
 \Leftrightarrow \left| {x - 3} \right| = 4\\
 \Leftrightarrow \left[ \begin{array}{l}
x - 3 = 4\\
x - 3 =  - 4
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 7\\
x =  - 1
\end{array} \right.\\
b,\\
DKXD:\,\,\,{x^2} + 3 \ge 0,\,\,\,\forall x\\
\sqrt {9{x^2} + 27}  + \sqrt {25{x^2} + 75}  - \sqrt {49{x^2} + 147}  = 3\\
 \Leftrightarrow \sqrt {9.\left( {{x^2} + 3} \right)}  + \sqrt {25.\left( {{x^2} + 3} \right)}  - \sqrt {49.\left( {{x^2} + 3} \right)}  = 3\\
 \Leftrightarrow \sqrt {{3^2}.\left( {{x^2} + 3} \right)}  + \sqrt {{5^2}.\left( {{x^2} + 3} \right)}  - \sqrt {{7^2}.\left( {{x^2} + 3} \right)}  = 3\\
 \Leftrightarrow 3.\sqrt {{x^2} + 3}  + 5.\sqrt {{x^2} + 3}  - 7.\sqrt {{x^2} + 3}  = 3\\
 \Leftrightarrow \sqrt {{x^2} + 3}  = 3\\
 \Leftrightarrow {x^2} + 3 = {3^2}\\
 \Leftrightarrow {x^2} + 3 = 9\\
 \Leftrightarrow {x^2} = 6\\
 \Leftrightarrow x =  \pm \sqrt 6 \\
c,\\
DKXD:\,\,\,x \ge 2\\
\sqrt {x + 1}  + \sqrt {x - 2}  + 2\sqrt {{x^2} - x - 2}  = 13 - 2x\\
 \Leftrightarrow \sqrt {x + 1}  + \sqrt {x - 2}  + 2\sqrt {{x^2} - x - 2}  + 2x - 13 = 0\\
 \Leftrightarrow \left( {\sqrt {x + 1}  - 2} \right) + \left( {\sqrt {x - 2}  - 1} \right) + 2.\left( {\sqrt {{x^2} - x - 2}  - 2} \right) + \left( {2x - 6} \right) = 0\\
 \Leftrightarrow \dfrac{{\left( {\sqrt {x + 1}  - 2} \right)\left( {\sqrt {x + 1}  + 2} \right)}}{{\sqrt {x + 1}  + 2}} + \dfrac{{\left( {\sqrt {x - 2}  - 1} \right)\left( {\sqrt x  - 2} \right) + 1}}{{\sqrt {x - 2}  + 1}}\\
 + 2.\dfrac{{\left( {\sqrt {{x^2} - x - 2}  - 2} \right)\left( {\sqrt {{x^2} - x - 2}  + 2} \right)}}{{\sqrt {{x^2} - x - 2}  + 2}} + 2.\left( {x - 3} \right) = 0\\
 \Leftrightarrow \dfrac{{{{\sqrt {x + 1} }^2} - {2^2}}}{{\sqrt {x + 1}  + 2}} + \dfrac{{{{\sqrt {x - 2} }^2} - {1^2}}}{{\sqrt {x - 2}  + 1}} + 2.\dfrac{{{{\sqrt {{x^2} - x - 2} }^2} - {2^2}}}{{\sqrt {{x^2} - x - 2}  + 2}} + 2.\left( {x - 3} \right) = 0\\
 \Leftrightarrow \dfrac{{x + 1 - 4}}{{\sqrt {x + 1}  + 2}} + \dfrac{{x - 2 - 1}}{{\sqrt {x - 2}  + 1}} + 2.\dfrac{{{x^2} - x - 2 - 4}}{{\sqrt {{x^2} - x - 2}  + 2}} + 2.\left( {x - 3} \right) = 0\\
 \Leftrightarrow \dfrac{{x - 3}}{{\sqrt {x + 1}  + 2}} + \dfrac{{x - 3}}{{\sqrt {x - 2}  + 1}} + 2.\dfrac{{{x^2} - x - 6}}{{\sqrt {{x^2} - x - 2}  + 2}} + 2.\left( {x - 3} \right) = 0\\
 \Leftrightarrow \dfrac{{x - 3}}{{\sqrt {x + 1}  + 2}} + \dfrac{{x - 3}}{{\sqrt {x - 2}  + 1}} + 2.\dfrac{{\left( {x - 3} \right)\left( {x + 2} \right)}}{{\sqrt {{x^2} - x - 2}  + 2}} + 2.\left( {x - 3} \right) = 0\\
 \Leftrightarrow \left( {x - 3} \right).\left[ {\dfrac{1}{{\sqrt {x + 1}  + 2}} + \dfrac{1}{{\sqrt {x - 2}  + 1}} + 2.\dfrac{{x + 2}}{{\sqrt {{x^2} - x - 2}  + 2}} + 2} \right] = 0\\
x \ge 2 \Rightarrow \dfrac{1}{{\sqrt {x + 1}  + 2}} + \dfrac{1}{{\sqrt {x - 2}  + 1}} + 2.\dfrac{{x + 2}}{{\sqrt {{x^2} - x - 2}  + 2}} + 2 > 0\\
 \Rightarrow x - 3 = 0\\
 \Leftrightarrow x = 3
\end{array}\)