Đáp án:
$\begin{array}{l}
Dkxd:x \ge 2;x\# 5\\
A = \dfrac{{x - 5}}{{\sqrt {x - 2} - 1}}\\
= \dfrac{{9 - 4\sqrt 3 - 5}}{{\sqrt {9 - 4\sqrt 3 - 2} - 1}}\\
= \dfrac{{4 - 4\sqrt 3 }}{{\sqrt {7 - 4\sqrt 3 } - 1}}\\
= \dfrac{{4 - 4\sqrt 3 }}{{\sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} - 1}}\\
= \dfrac{{4.\left( {1 - \sqrt 3 } \right)}}{{2 - \sqrt 3 - 1}}\\
= \dfrac{{4\left( {1 - \sqrt 3 } \right)}}{{1 - \sqrt 3 }}\\
= 4\\
B = \sqrt {4 + {{\left( {1 + 6x + 9{x^2}} \right)}^2}} \\
= \sqrt {4 + {{\left( {3x + 1} \right)}^2}} \\
= \sqrt {4 + {{\left( {3.\left( { - \sqrt 2 } \right) + 1} \right)}^2}} \\
= \sqrt {4 + 12 - 6\sqrt 2 + 1} \\
= \sqrt {17 - 6\sqrt 2 } \\
= \sqrt {9 - 2.3.2\sqrt 2 + 8} \\
= \sqrt {{{\left( {3 - 2\sqrt 2 } \right)}^2}} \\
= 3 - 2\sqrt 2 \\
C = 4x - 7 + \dfrac{{\sqrt {{x^3} + 3{x^2}} }}{{\sqrt {x + 2} }}\\
= 4.\left( { - \sqrt 2 } \right) - 7 + \dfrac{{\sqrt {{{\left( { - \sqrt 2 } \right)}^3} + 3.{{\left( { - \sqrt 2 } \right)}^2}} }}{{\sqrt { - \sqrt 2 + 2} }}\\
= - 4\sqrt 2 - 7 + \dfrac{{\sqrt 4 }}{{\sqrt {2 - \sqrt 2 } }}\\
= - 4\sqrt 2 - 7 + \dfrac{{2\sqrt {2 - \sqrt 2 } }}{{2 - \sqrt 2 }}\\
= - 4\sqrt 2 - 7 + \dfrac{{2\sqrt {2 - \sqrt 2 } \left( {2 + \sqrt 2 } \right)}}{2}\\
= - 4\sqrt 2 - 7 + \sqrt {2 - \sqrt 2 } \left( {2 + \sqrt 2 } \right)\\
D = \dfrac{1}{{\sqrt {x + 2\sqrt {x - 1} } }} + \dfrac{1}{{\sqrt {x - 2\sqrt {x - 1} } }}\\
= \dfrac{1}{{\sqrt {{{\left( {\sqrt {x - 1} + 1} \right)}^2}} }} + \dfrac{1}{{\sqrt {{{\left( {\sqrt {x - 1} - 1} \right)}^2}} }}\\
= \dfrac{1}{{\sqrt {x - 1} + 1}} + \dfrac{1}{{\sqrt {x - 1} - 1}}\\
= \dfrac{{\sqrt {x - 1} - 1 + \sqrt {x - 1} + 1}}{{x - 1 + 1}}\\
= \dfrac{{2\sqrt {x - 1} }}{x}\\
= \dfrac{{2\sqrt {10 - 1} }}{{10}}\\
= \dfrac{3}{5}
\end{array}$
