Đáp án:

$\begin{array}{l}
1)\\
a)x + 3 + \sqrt {{x^2} - 6x + 9} \left( {x \le 3} \right)\\
 = x + 3 + \sqrt {{{\left( {x - 3} \right)}^2}} \left( {x \le 3} \right)\\
 = x + 3 + \left| {x - 3} \right|\left( {x \le 3} \right)\\
 = x + 3 + 3 - x\\
 = 6\\
b)\sqrt {{x^2} + 4x + 4}  - \sqrt {{x^2}} \left( { - 2 \le x \le 0} \right)\\
 = \sqrt {{{\left( {x + 2} \right)}^2}}  - \sqrt {{x^2}} \\
 = \left| {x + 2} \right| - \left| x \right|\\
 = x + 2 - \left( { - x} \right)\\
 = 2x + 2\\
c)\dfrac{{\sqrt {{x^2} - 2x + 1} }}{{x - 1}}\left( {x > 1} \right)\\
 = \dfrac{{\sqrt {{{\left( {x - 1} \right)}^2}} }}{{x - 1}}\\
 = \dfrac{{x - 1}}{{x - 1}}\\
 = 1\\
d)\left| {x - 2} \right| + \dfrac{{\sqrt {{x^2} - 4x + 4} }}{{x - 2}}\left( {x < 2} \right)\\
 = 2 - x + \dfrac{{\sqrt {{{\left( {x - 2} \right)}^2}} }}{{x - 2}}\\
 = 2 - x + \dfrac{{2 - x}}{{x - 2}}\\
 = 2 - x - 1\\
 = 1 - x\\
B2)a)\sqrt {1 - 4a + 4{a^2}}  - 2a\\
 = \sqrt {{{\left( {2a - 1} \right)}^2}}  - 2a\\
 = \left| {2a - 1} \right| - 2a\\
 = \left[ \begin{array}{l}
2a - 1 - 2a =  - 1\left( {a \ge \dfrac{1}{2}} \right)\\
1 - 2a - 2a = 1 - 4a\left( {a < \dfrac{1}{2}} \right)
\end{array} \right.\\
b)x - 2y - \sqrt {{x^2} - 4xy + 4{y^2}} \\
 = x - 2y - \sqrt {{{\left( {x - 2y} \right)}^2}} \\
 = x - 2y - \left| {x - 2y} \right|\\
 = \left[ \begin{array}{l}
x - 2y - \left( {x - 2y} \right) = 0\left( {khi:x \ge 2y} \right)\\
x - 2y + x - 2y = 2x - 4y\left( {khi:x < 2y} \right)
\end{array} \right.\\
c){x^2} + \sqrt {{x^4} - 8{x^2} + 16} \\
 = {x^2} + \sqrt {{{\left( {{x^2} - 4} \right)}^2}} \\
 = {x^2} + \left| {{x^2} - 4} \right|\\
 = \left[ \begin{array}{l}
{x^2} + {x^2} - 4 = 2{x^2} - 4\left( {khi:{x^2} \ge 4} \right)\\
{x^2} - x + 4 = 4\left( {khi:{x^2} < 4} \right)
\end{array} \right.\\
d)2x - 1 - \dfrac{{\sqrt {{x^2} - 10x + 25} }}{{x - 5}}\\
 = 2x - 1 - \dfrac{{\sqrt {{{\left( {x - 5} \right)}^2}} }}{{x - 5}}\\
 = 2x - 1 - \dfrac{{\left| {x - 5} \right|}}{{x - 5}}\\
 = \left[ \begin{array}{l}
2x - 1 - \dfrac{{x - 5}}{{x - 5}} = 2x - 1 - 1 = 2x - 2\left( {khi:x > 5} \right)\\
2x - 1 + \dfrac{{x - 5}}{{x - 5}} = 2x - 1 + 1 = 2x\left( {khi:x < 5} \right)
\end{array} \right.\\
e)\dfrac{{\sqrt {{x^4} - 4{x^2} + 4} }}{{{x^2} - 2}}\\
 = \dfrac{{\sqrt {{{\left( {{x^2} - 2} \right)}^2}} }}{{{x^2} - 2}} = \dfrac{{\left| {{x^2} - 2} \right|}}{{{x^2} - 2}}\\
 = \left[ \begin{array}{l}
\dfrac{{{x^2} - 2}}{{{x^2} - 2}} = 1\left( {khi:{x^2} > 2} \right)\\
\dfrac{{ - {x^2} + 2}}{{{x^2} - 2}} =  - 1\left( {khi:{x^2} < 2} \right)
\end{array} \right.\\
f)\sqrt {{{\left( {x - 4} \right)}^2}}  + \dfrac{{x - 4}}{{\sqrt {{x^2} - 8x + 16} }}\\
 = \left| {x - 4} \right| + \dfrac{{x - 4}}{{\left| {x - 4} \right|}}\\
 = \left[ \begin{array}{l}
x - 4 + \dfrac{{x - 4}}{{x - 4}} = x - 3\left( {khi:x > 4} \right)\\
4 - x + \dfrac{{x - 4}}{{4 - x}} = 3 - x\left( {khi:x < 4} \right)
\end{array} \right.
\end{array}$