Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
3x.\left( {x - 2{x^2} + 5} \right)\\
 = 3x.x - 3x.2{x^2} + 3x.5\\
 = 3{x^2} - 6{x^3} + 15x\\
 =  - 6{x^3} + 3{x^2} + 15x\\
b,\\
\left( {3x + 5} \right).\left( {{x^2} - 3} \right)\\
 = 3x.{x^2} - 3x.3 + 5.{x^2} - 5.3\\
 = 3{x^3} - 9x + 5{x^2} - 15\\
 = 3{x^3} + 5{x^2} - 9x - 15\\
c,\\
\left( {6{x^3}{y^3} - 18{x^2}{y^3} - 3x{y^2}} \right):3x{y^2}\\
 = \left( {3x{y^2}.2{x^2}y - 3x{y^2}.6xy - 3x{y^2}.1} \right):3x{y^2}\\
 = \left[ {3x{y^2}.\left( {2{x^2}y - 6xy - 1} \right)} \right]:3x{y^2}\\
 = 2{x^2}y - 6xy - 1\\
2,\\
a,\\
5xy - 15x{y^2} = 5xy - 5xy.3y = 5xy\left( {1 - 3y} \right)\\
b,\\
{x^2} - xy - 5x + 5y = \left( {{x^2} - xy} \right) + \left( { - 5x + 5y} \right)\\
 = x.\left( {x - y} \right) - 5.\left( {x - y} \right) = \left( {x - y} \right)\left( {x - 5} \right)\\
c,\\
6{x^2} - 7x + 2 = \left( {6{x^2} - 4x} \right) + \left( { - 3x + 2} \right)\\
 = 2x.\left( {3x - 2} \right) - \left( {3x - 2} \right) = \left( {3x - 2} \right)\left( {2x - 1} \right)\\
3,\\
A =  - 5{x^2} - 5x - {y^2} + 4y + 2xy - 15\\
 = \left( { - {x^2} + 2xy - {y^2}} \right) + \left( { - 4x + 4y} \right) + \left( { - 4{x^2} - x - \dfrac{1}{{16}}} \right) - \dfrac{{239}}{{16}}\\
 =  - \left( {{x^2} - 2xy + {y^2}} \right) - 4.\left( {x - y} \right) - \left( {4{x^2} + x + \dfrac{1}{{16}}} \right) - \dfrac{{239}}{{16}}\\
 =  - {\left( {x - y} \right)^2} - 4.\left( {x - y} \right) - \left[ {{{\left( {2x} \right)}^2} + 2.2x.\dfrac{1}{4} + {{\left( {\dfrac{1}{4}} \right)}^2}} \right] - \dfrac{{239}}{{16}}\\
 = \left[ { - {{\left( {x - y} \right)}^2} - 4\left( {x - y} \right) - 4} \right] - {\left( {2x + \dfrac{1}{4}} \right)^2} - \dfrac{{175}}{{16}}\\
 =  - \left[ {{{\left( {x - y} \right)}^2} + 4\left( {x - y} \right) + 4} \right] - {\left( {2x + \dfrac{1}{4}} \right)^2} - \dfrac{{175}}{{16}}\\
 =  - {\left[ {\left( {x - y} \right) + 2} \right]^2} - {\left( {2x + \dfrac{1}{4}} \right)^2} - \dfrac{{175}}{{16}}\\
 =  - \left[ {{{\left( {x - y + 2} \right)}^2} + {{\left( {2x + \dfrac{1}{4}} \right)}^2} + \dfrac{{175}}{{16}}} \right]\\
{\left( {x - y + 2} \right)^2} + {\left( {2x + \dfrac{1}{4}} \right)^2} + \dfrac{{175}}{{16}} \ge \dfrac{{175}}{{16}},\,\,\,\,\forall x,y\\
 \Rightarrow A =  - \left[ {{{\left( {x - y + 2} \right)}^2} + {{\left( {2x + \dfrac{1}{4}} \right)}^2} + \dfrac{{175}}{{16}}} \right] \le  - \dfrac{{175}}{{16}}\\
 \Rightarrow {A_{\max }} =  - \dfrac{{175}}{{16}} \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - y + 2} \right)^2} = 0\\
{\left( {2x + \dfrac{1}{4}} \right)^2} = 0
\end{array} \right.\\
 \Leftrightarrow \left\{ \begin{array}{l}
x - y + 2 = 0\\
2x + \dfrac{1}{4} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x =  - \dfrac{1}{8}\\
y = \dfrac{{15}}{8}
\end{array} \right.
\end{array}\)