Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
2{x^2} - 2xy - 3x + 3y\\
 = \left( {2{x^2} - 2xy} \right) - \left( {3x - 3y} \right)\\
 = 2x.\left( {x - y} \right) - 3.\left( {x - y} \right)\\
 = \left( {x - y} \right)\left( {2x - 3} \right)\\
2,\\
 - {x^2} - {y^2} + 2xy + 16\\
 = 16 - \left( {{x^2} - 2xy + {y^2}} \right)\\
 = {4^2} - {\left( {x - y} \right)^2}\\
 = \left[ {4 - \left( {x - y} \right)} \right].\left[ {4 + \left( {x - y} \right)} \right]\\
 = \left( { - x + y + 4} \right)\left( {x - y + 4} \right)\\
3,\\
{y^2} - {x^2} + 2yz + {z^2}\\
 = \left( {{y^2} + 2yz + {z^2}} \right) - {x^2}\\
 = {\left( {y + z} \right)^2} - {x^2}\\
 = \left( {y + z - x} \right)\left( {y + z + x} \right)\\
4,\\
3{x^2} - 6xy + 3{y^2} - 12{z^2}\\
 = 3.\left[ {\left( {{x^2} - 2xy + {y^2}} \right) - 4{z^2}} \right]\\
 = 3.\left[ {{{\left( {x - y} \right)}^2} - {{\left( {2z} \right)}^2}} \right]\\
 = 3.\left( {x - y + 2z} \right)\left( {x - y - 2z} \right)\\
5,\\
{x^4} - 2{x^3} + 2x - 1\\
 = \left( {{x^4} - 2{x^3} + {x^2}} \right) - \left( {{x^2} - 2x + 1} \right)\\
 = {x^2}\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)\\
 = \left( {{x^2} - 2x + 1} \right)\left( {{x^2} - 1} \right)\\
 = {\left( {x - 1} \right)^2}\left( {x - 1} \right)\left( {x + 1} \right)\\
 = {\left( {x - 1} \right)^3}\left( {x + 1} \right)\\
6,\\
{a^6} - {a^4} + 2{a^3} + 2{a^2}\\
 = {a^2}\left( {{a^4} - {a^2} + 2a + 2} \right)\\
 = {a^2}.\left[ {\left( {{a^4} - {a^2}} \right) + \left( {2a + 2} \right)} \right]\\
 = {a^2}.\left[ {{a^2}\left( {{a^2} - 1} \right) + 2.\left( {a + 1} \right)} \right]\\
 = {a^2}.\left[ {{a^2}\left( {a - 1} \right)\left( {a + 1} \right) + 2.\left( {a + 1} \right)} \right]\\
 = {a^2}.\left( {a + 1} \right).\left[ {{a^2}\left( {a + 1} \right) + 2} \right]\\
 = {a^2}\left( {a + 1} \right)\left( {{a^3} + {a^2} + 2} \right)\\
7,\\
{x^4} + {x^3} + 2{x^2} + x + 1\\
 = \left( {{x^4} + {x^3} + {x^2}} \right) + \left( {{x^2} + x + 1} \right)\\
 = {x^2}\left( {{x^2} + x + 1} \right) + \left( {{x^2} + x + 1} \right)\\
 = \left( {{x^2} + x + 1} \right)\left( {{x^2} + 1} \right)\\
8,\\
{x^2}y + x{y^2} + {x^2}z + {y^2}z + 2xyz\\
 = \left( {{x^2}y + x{y^2}} \right) + \left( {{x^2}z + {y^2}z + 2xyz} \right)\\
 = xy\left( {x + y} \right) + z.\left( {{x^2} + {y^2} + 2xy} \right)\\
 = xy\left( {x + y} \right) + z.{\left( {x + y} \right)^2}\\
 = \left( {x + y} \right).\left[ {xy + z.\left( {x + y} \right)} \right]\\
 = \left( {x + y} \right).\left( {xy + yz + zx} \right)\\
9,\\
{x^4} + 2{x^3} + 2{x^2} + 2x + 1\\
 = \left( {{x^4} + 2{x^3} + {x^2}} \right) + \left( {{x^2} + 2x + 1} \right)\\
 = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\
 = \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right)\\
 = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right)\\
10,\,\,\,\,\,{x^5} + {x^4} + {x^3} + {x^2} + x + 1\\
 = \left( {{x^5} + {x^4}} \right) + \left( {{x^3} + {x^2}} \right) + \left( {x + 1} \right)\\
 = {x^4}\left( {x + 1} \right) + {x^2}\left( {x + 1} \right) + \left( {x + 1} \right)\\
 = \left( {x + 1} \right)\left( {{x^4} + {x^2} + 1} \right)\\
 = \left( {x + 1} \right).\left[ {\left( {{x^4} + 2{x^2} + 1} \right) - {x^2}} \right]\\
 = \left( {x + 1} \right).\left[ {{{\left( {{x^2} + 1} \right)}^2} - {x^2}} \right]\\
 = \left( {x + 1} \right).\left( {{x^2} + 1 - x} \right)\left( {{x^2} + 1 + x} \right)
\end{array}\)