Đáp án:

\(\begin{array}{l}
1,\\
a,\,\,\,{\left( {5x - 6} \right)^2}\\
b,\,\,\,4.{\left( {x - 2} \right)^2}\\
c,\,\,\,{\left( {3x - 1} \right)^2}\\
d,\,\,\,16\\
e,\,\,\,{\left( {2x + 5} \right)^2}\\
2,\\
a,\\
{A_{\min }} = 1 \Leftrightarrow x =  - 3\\
b,\\
{B_{\min }} = 4 \Leftrightarrow x =  - 2\\
3,\\
x = 2
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
a,\\
{\left( {2x - 3} \right)^2} + 2.\left( {2x - 3} \right).\left( {3x - 3} \right) + {\left( {3x - 3} \right)^2}\\
 = {\left[ {\left( {2x - 3} \right) + \left( {3x - 3} \right)} \right]^2}\\
 = {\left( {2x - 3 + 3x - 3} \right)^2}\\
 = {\left( {5x - 6} \right)^2}\\
b,\\
4{x^2} - 16x + 16\\
 = 4.\left( {{x^2} - 4x + 4} \right)\\
 = 4.\left( {{x^2} - 2.x.2 + {2^2}} \right)\\
 = 4.{\left( {x - 2} \right)^2}\\
c,\\
9{x^2} - 6x + 1\\
 = {\left( {3x} \right)^2} - 2.3x.1 + {1^2}\\
 = {\left( {3x - 1} \right)^2}\\
d,\\
{\left( {4x + 2} \right)^2} + {\left( {4x - 2} \right)^2} - 2.\left( {4x + 2} \right)\left( {4x - 2} \right)\\
 = {\left( {4x + 2} \right)^2} - 2.\left( {4x + 2} \right).\left( {4x - 2} \right) + {\left( {4x - 2} \right)^2}\\
 = {\left[ {\left( {4x + 2} \right) - \left( {4x - 2} \right)} \right]^2}\\
 = {\left( {4x + 2 - 4x + 2} \right)^2}\\
 = {4^2}\\
 = 16\\
e,\\
4{x^2} + 20x + 25\\
 = {\left( {2x} \right)^2} + 2.2x.5 + {5^2}\\
 = {\left( {2x + 5} \right)^2}\\
2,\\
a,\\
A = {x^2} + 6x + 10 = \left( {{x^2} + 6x + 9} \right) + 1\\
 = \left( {{x^2} + 2.x.3 + {3^2}} \right) + 1 = {\left( {x + 3} \right)^2} + 1\\
{\left( {x + 3} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow {\left( {x + 3} \right)^2} + 1 \ge 1,\,\,\,\forall x\\
 \Rightarrow A \ge 1,\,\,\,\forall x\\
 \Rightarrow {A_{\min }} = 1 \Leftrightarrow {\left( {x + 3} \right)^2} = 0 \Leftrightarrow x + 3 = 0 \Leftrightarrow x =  - 3\\
b,\\
B = 4{x^2} + 16x + 20 = \left( {4{x^2} + 16x + 16} \right) + 4\\
 = \left[ {{{\left( {2x} \right)}^2} + 2.2x.4 + {4^2}} \right] + 4 = {\left( {2x + 4} \right)^2} + 4\\
{\left( {2x + 4} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow {\left( {2x + 4} \right)^2} + 4 \ge 4,\,\,\,\,\forall x\\
 \Rightarrow B \ge 4,\,\,\,\forall x\\
 \Rightarrow {B_{\min }} = 4 \Leftrightarrow {\left( {2x + 4} \right)^2} = 0 \Leftrightarrow 2x + 4 = 0 \Leftrightarrow x =  - 2\\
3,\\
\left| {x - 3} \right| = 2x - 3\\
 \Leftrightarrow \left\{ \begin{array}{l}
2x - 3 \ge 0\\
\left[ \begin{array}{l}
x - 3 = 2x - 3\\
x - 3 = 3 - 2x
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{3}{2}\\
\left[ \begin{array}{l}
x - 2x =  - 3 + 3\\
x + 2x = 3 + 3
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{3}{2}\\
\left[ \begin{array}{l}
 - x = 0\\
3x = 6
\end{array} \right.
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x \ge \dfrac{3}{2}\\
\left[ \begin{array}{l}
x = 0\\
x = 2
\end{array} \right.
\end{array} \right.\\
 \Rightarrow x = 2
\end{array}\)