Giúp mik bài 1,2 với

Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
a,\\
2x.\left( {{x^2} - 7x - 3} \right) = 2x.{x^2} - 2x.7x - 2x.3 = 2{x^3} - 14{x^2} - 6x\\
b,\\
\left( {5x - 3} \right)\left( {3 + 5x} \right) = \left( {5x - 3} \right)\left( {5x + 3} \right) = {\left( {5x} \right)^2} - {3^2} = 25{x^2} - 9\\
c,\\
\left( {2{x^4} - {x^3} + 3{x^2}} \right):\left( { - \frac{1}{3}{x^2}} \right)\\
 = \left[ {{x^2}\left( {2{x^2} - x + 3} \right)} \right]:\left( { - \frac{1}{3}{x^2}} \right)\\
 = \left( {2{x^2} - x + 3} \right):\left( { - \frac{1}{3}} \right)\\
 = \left( {2{x^2} - x + 3} \right).\left( { - 3} \right)\\
 =  - 6{x^2} + 3x - 9\\
d,\\
\left( {x + 2} \right)\left( {{x^2} - 2x + 4} \right)\\
 = \left( {x + 2} \right).\left( {{x^2} - x.2 + {2^2}} \right)\\
 = {x^3} + {2^3}\\
 = {x^3} + 8\\
2,\\
a,\\
12{x^2}y - 18x{y^2} - 30{y^2}\\
 = 6y.2{x^2} - 6y.3xy - 6y.5y\\
 = 6y.\left( {2{x^2} - 3xy - 5y} \right)\\
b,\\
3x - {x^2} - {y^2} + 3y - 2xy\\
 = \left( {3x + 3y} \right) + \left( { - {x^2} - {y^2} - 2xy} \right)\\
 = 3.\left( {x + y} \right) - \left( {{x^2} + 2xy + {y^2}} \right)\\
 = 3.\left( {x + y} \right) - {\left( {x + y} \right)^2}\\
 = \left( {x + y} \right).\left[ {3 - \left( {x + y} \right)} \right]\\
 = \left( {x + y} \right)\left( {3 - x - y} \right)\\
c,\\
\left( {{x^2} + x + 1} \right)\left( {{x^2} + x + 2} \right) - 12\\
 = \left( {{x^2} + x + 1} \right)\left[ {\left( {{x^2} + x + 1} \right) + 1} \right] - 12\\
 = {\left( {{x^2} + x + 1} \right)^2} + \left( {{x^2} + x + 1} \right) - 12\\
 = \left[ {{{\left( {{x^2} + x + 1} \right)}^2} - 4.\left( {{x^2} + x + 1} \right)} \right] + \left[ {3.\left( {{x^2} + x + 1} \right) - 12} \right]\\
 = \left( {{x^2} + x + 1} \right)\left[ {\left( {{x^2} + x + 1} \right) - 4} \right] + 3.\left[ {\left( {{x^2} + x + 1} \right) - 4} \right]\\
 = \left[ {\left( {{x^2} + x + 1} \right) - 4} \right].\left[ {\left( {{x^2} + x + 1} \right) + 3} \right]\\
 = \left( {{x^2} + x - 3} \right)\left( {{x^2} + x + 4} \right)
\end{array}\)