Đáp án:
$\begin{array}{l}
1)a)\sqrt {{x^2} - 12x + 36} = 81\\
\Leftrightarrow \sqrt {{{\left( {x - 6} \right)}^2}} = 81\\
\Leftrightarrow \left| {x - 6} \right| = 81\\
\Leftrightarrow \left[ \begin{array}{l}
x - 6 = 81\\
x - 6 = - 81
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 87\\
x = - 75
\end{array} \right.\\
Vậy\,x = 87;x = - 75\\
d)\sqrt {{x^2} - 6x + 9} = \sqrt {4{x^2}} \\
\Leftrightarrow \sqrt {{{\left( {x - 3} \right)}^2}} = \sqrt {{{\left( {2x} \right)}^2}} \\
\Leftrightarrow \left[ \begin{array}{l}
x - 3 = 2x\\
x - 3 = - 2x
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = - 3\\
x = 1
\end{array} \right.\\
Vậy\,x = - 3;x = 1\\
B2)\\
a)Dkxd:\left\{ \begin{array}{l}
x > 0\\
x \ne 4
\end{array} \right.\\
b)A = \left( {\dfrac{{\sqrt x - 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}} + \dfrac{3}{{\sqrt x - 2}}} \right):\left( {\dfrac{{\sqrt x + 2}}{{\sqrt x }} - \dfrac{{\sqrt x }}{{\sqrt x - 2}}} \right)\\
= \dfrac{{\sqrt x - 4 + 3\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}:\dfrac{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 2} \right) - \sqrt x .\sqrt x }}{{\sqrt x \left( {\sqrt x - 2} \right)}}\\
= \dfrac{{4\sqrt x - 4}}{{\sqrt x \left( {\sqrt x - 2} \right)}}.\dfrac{{\sqrt x \left( {\sqrt x - 2} \right)}}{{x - 4 - x}}\\
= \dfrac{{4\left( {\sqrt x - 1} \right)}}{{ - 4}}\\
= 1 - \sqrt x \\
c)\dfrac{7}{A} = \dfrac{7}{{1 - \sqrt x }} \in Z\\
\Leftrightarrow 1 - \sqrt x \in \left\{ { - 7; - 1;1;7} \right\}\\
\Leftrightarrow \sqrt x \in \left\{ {8;2;0; - 6} \right\}\\
Do:\sqrt x > 0;\sqrt x \ne 2\left( {khi:x > 0;x \ne 4} \right)\\
\Leftrightarrow \sqrt x = 8\\
\Leftrightarrow x = 64\left( {tm} \right)\\
Vậy\,x = 64
\end{array}$
