Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
a,\\
{\left( {3x - 5} \right)^2} - 2.\left( {3x - 5} \right).\left( {3x + 5} \right) + {\left( {3x + 5} \right)^2}\\
 = {\left[ {\left( {3x - 5} \right) - \left( {3x + 5} \right)} \right]^2}\\
 = {\left( { - 10} \right)^2}\\
 = 100\\
b,\\
3.\left( {2x - 1} \right) - 5.\left( {x - 3} \right) + \left( {3x - 4} \right) - 49x\\
 = \left( {6x - 3} \right) - \left( {5x - 15} \right) + 3x - 4 - 49x\\
 = 6x - 3 - 5x + 15 + 3x - 4 - 49x\\
 =  - 45x + 8\\
2,\\
a,\\
{x^2} - 2xy - {z^2} + {y^2}\\
 = \left( {{x^2} - 2xy + {y^2}} \right) - {z^2}\\
 = {\left( {x - y} \right)^2} - {z^2}\\
 = \left( {x - y - z} \right)\left( {x - y + z} \right)\\
b,\\
{x^2} - 2x - 15\\
 = \left( {{x^2} - 5x} \right) + \left( {3x - 15} \right)\\
 = x\left( {x - 5} \right) + 3.\left( {x - 5} \right)\\
 = \left( {x - 5} \right)\left( {x + 3} \right)\\
3,\\
A = 2.{\left( {x - 1} \right)^2} + 2.{\left( {x + 1} \right)^2} + {\left( {y - 1} \right)^2} + 2020\\
 = 2.\left( {{x^2} - 2x + 1} \right) + 2.\left( {{x^2} + 2x + 1} \right) + {\left( {y - 1} \right)^2} + 2020\\
 = 4{x^2} + 4 + {\left( {y - 1} \right)^2} + 2020\\
 = 4{x^2} + {\left( {y - 1} \right)^2} + 2024 \ge 2024,\,\,\,\forall x,y\\
 \Rightarrow {A_{\min }} = 2024 \Leftrightarrow \left\{ \begin{array}{l}
{x^2} = 0\\
{\left( {y - 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 0\\
y = 1
\end{array} \right.\\
B = 5{x^2} + {y^2} - 4xy - 2y + 2021\\
 = \left( {4{x^2} - 4xy + {y^2}} \right) + \left( {4x - 2y} \right) + \left( {{x^2} - 4x + 4} \right) + 2017\\
 = {\left( {2x - y} \right)^2} + 2.\left( {2x - y} \right) + {\left( {x - 2} \right)^2} + 2017\\
 = \left[ {{{\left( {2x - y} \right)}^2} + 2.\left( {2x - y} \right) + 1} \right] + {\left( {x - 2} \right)^2} + 2016\\
 = {\left( {2x - y + 1} \right)^2} + {\left( {x - 2} \right)^2} + 2016 \ge 2016,\,\,\,\forall x,y\\
 \Rightarrow {B_{\min }} = 2016 \Leftrightarrow \left\{ \begin{array}{l}
2x - y + 1 = 0\\
x - 2 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
y = 5
\end{array} \right.\\
C = 2{x^2} + 2x + {y^2} - 2xy + 2017\\
 = \left( {{x^2} - 2xy + {y^2}} \right) + \left( {{x^2} + 2x + 1} \right) + 2016\\
 = {\left( {x - y} \right)^2} + {\left( {x + 1} \right)^2} + 2016 \ge 2016,\,\,\,\forall x,y\\
 \Rightarrow {C_{\min }} = 2016 \Rightarrow \left\{ \begin{array}{l}
{\left( {x - y} \right)^2} = 0\\
{\left( {x + 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow x = y =  - 1
\end{array}\)