Đáp án:
$\begin{array}{l}
9)x + 2y + \sqrt {{x^2} - 4xy + 4{y^2}} \\
= x + 2y + \sqrt {{{\left( {x - 2y} \right)}^2}} \\
= x + 2y + \left| {x - 2y} \right|\\
= \left[ \begin{array}{l}
x + 2y + x - 2y = 2x\left( {khi:x \ge 2y} \right)\\
x + 2y + 2y - x = 4y\left( {khi:x < 2y} \right)
\end{array} \right.\\
11)\dfrac{{\sqrt {{x^2} + 6x + 9} }}{{x + 3}}\\
= \dfrac{{\sqrt {{{\left( {x + 3} \right)}^2}} }}{{x + 3}}\\
= \dfrac{{\left| {x + 3} \right|}}{{x + 3}}\\
= \left[ \begin{array}{l}
= 1\left( {khi:x > - 3} \right)\\
= - 1\left( {khi:x < - 3} \right)
\end{array} \right.\\
12)\\
1 + \dfrac{{\sqrt {{{\left( {x - 1} \right)}^2}} }}{{x - 1}} = 1 + \dfrac{{\left| {x - 1} \right|}}{{x - 1}}\\
= \left[ \begin{array}{l}
1 + 1 = 2\left( {khi:x > 1} \right)\\
1 - 1 = 0\left( {khi:x < 1} \right)
\end{array} \right.\\
13)\dfrac{{\sqrt {{x^2} + 10x + 25} }}{{{x^2} - 25}}\\
= \dfrac{{\sqrt {{{\left( {x + 5} \right)}^2}} }}{{\left( {x + 5} \right)\left( {x - 5} \right)}}\\
= \dfrac{{\left| {x + 5} \right|}}{{\left( {x + 5} \right)\left( {x - 5} \right)}}\\
= \left[ \begin{array}{l}
\dfrac{1}{{x - 5}}\left( {khi:x > - 5} \right)\\
\dfrac{1}{{5 - x}}\left( {khi:x < - 5} \right)
\end{array} \right.\\
14)\sqrt {{{\left( {x - 2} \right)}^2}} + \dfrac{{x - 2}}{{\sqrt {{{\left( {x - 2} \right)}^2}} }}\\
= \left| {x - 2} \right| + \dfrac{{x - 2}}{{\left| {x - 2} \right|}}\\
= \left[ \begin{array}{l}
x - 2 + 1 = x - 1\left( {khi:x > 2} \right)\\
2 - x - 1 = 1 - x\left( {khi:x < 2} \right)
\end{array} \right.
\end{array}$
