Đáp án:

\(\begin{array}{l}
1,\\
a,\\
15{x^5} - 10{x^4} + 5{x^3} + 10{x^2}\\
b,\\
 - 2{a^5}{x^4} + 10{x^2}{a^3} - 6{a^4}x\\
c,\\
6{x^4} - 2{x^3} - 15{x^2} + 23x - 6\\
d,\\
{a^5} - {b^5}\\
2,\\
a,\\
{a^4} - {a^2} + 2a - 1\\
b,\\
{a^6} - 64\\
c,\\
4 + 12y - 4{x^2}\\
d,\\
 - 2{x^3} + 6{x^2} + 4
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
1,\\
a,\\
\left( {3{x^3} - 2{x^2} + x + 2} \right).\left( {5{x^2}} \right)\\
 = 3{x^3}.5{x^2} - 2{x^2}.5{x^2} + x.5{x^2} + 2.5{x^2}\\
 = 15{x^5} - 10{x^4} + 5{x^3} + 10{x^2}\\
b,\\
\left( {{a^2}{x^3} - 5x + 3a} \right).\left( { - 2{a^3}x} \right)\\
 = {a^2}{x^3}.\left( { - 2{a^3}x} \right) - 5x.\left( { - 2{a^3}x} \right) + 3a.\left( { - 2{a^3}x} \right)\\
 =  - 2.{a^2}.{a^3}.{x^3}.x + 10.x.x.{a^3} - 6.a.{a^3}.x\\
 =  - 2{a^5}{x^4} + 10{x^2}{a^3} - 6{a^4}x\\
c,\\
\left( {3{x^2} + 5x - 2} \right)\left( {2{x^2} - 4x + 3} \right)\\
 = 3{x^2}.2{x^2} + 3{x^2}.\left( { - 4x} \right) + 3{x^2}.3 + 5x.2{x^2} + 5x.\left( { - 4x} \right) + 5x.3 - 2.2{x^2} - 2.\left( { - 4x} \right) - 2.3\\
 = 6{x^4} - 12{x^3} + 9{x^2} + 10{x^3} - 20{x^2} + 15x - 4{x^2} + 8x - 6\\
 = 6{x^4} + \left( { - 12{x^3} + 10{x^3}} \right) + \left( {9{x^2} - 20{x^2} - 4{x^2}} \right) + \left( {15x + 8x} \right) - 6\\
 = 6{x^4} - 2{x^3} - 15{x^2} + 23x - 6\\
d,\\
\left( {{a^4} + {a^3}b + {a^2}{b^2} + a{b^3} + {b^4}} \right)\left( {a - b} \right)\\
 = \left( {{a^4} + {a^3}b + {a^2}{b^2} + a{b^3} + {b^4}} \right).a - \left( {{a^4} + {a^3}b + {a^2}{b^2} + a{b^3} + {b^4}} \right).b\\
 = {a^5} + {a^4}b + {a^3}{b^2} + {a^2}{b^3} + a{b^4} - {a^4}b - {a^3}{b^2} - {a^2}{b^3} - a{b^4} - {b^5}\\
 = {a^5} + \left( {{a^4}b - {a^4}b} \right) + \left( {{a^3}{b^2} - {a^3}{b^2}} \right) + \left( {{a^2}{b^3} - {a^2}{b^3}} \right) + \left( {a{b^4} - a{b^4}} \right) - {b^5}\\
 = {a^5} - {b^5}\\
2,\\
a,\\
\left( {{a^2} + a - 1} \right).\left( {{a^2} - a + 1} \right)\\
 = \left[ {{a^2} + \left( {a - 1} \right)} \right].\left[ {{a^2} - \left( {a - 1} \right)} \right]\\
 = {\left( {{a^2}} \right)^2} - {\left( {a - 1} \right)^2}\\
 = {a^4} - \left( {{a^2} - 2.a.1 + {1^2}} \right)\\
 = {a^4} - \left( {{a^2} - 2a + 1} \right)\\
 = {a^4} - {a^2} + 2a - 1\\
b,\\
\left( {a + 2} \right)\left( {a - 2} \right)\left( {{a^2} + 2a + 4} \right)\left( {{a^2} - 2a + 4} \right)\\
 = \left[ {\left( {a + 2} \right)\left( {{a^2} - 2a + 4} \right)} \right].\left[ {\left( {a - 2} \right)\left( {{a^2} + 2a + 4} \right)} \right]\\
 = \left[ {\left( {a + 2} \right).\left( {{a^2} - a.2 + {2^2}} \right)} \right].\left[ {\left( {a - 2} \right).\left( {{a^2} + a.2 + {2^2}} \right)} \right]\\
 = \left( {{a^3} + {2^3}} \right).\left( {{a^3} - {2^3}} \right)\\
 = \left( {{a^3} + 8} \right)\left( {{a^3} - 8} \right)\\
 = {\left( {{a^3}} \right)^2} - {8^2}\\
 = {a^6} - 64\\
c,\\
{\left( {2 + 3y} \right)^2} - {\left( {2x - 3y} \right)^2} - 12xy\\
 = \left[ {{2^2} + 2.2.3y + {{\left( {3y} \right)}^2}} \right] - \left[ {{{\left( {2x} \right)}^2} - 2.2x.3y + {{\left( {3y} \right)}^2}} \right] - 12xy\\
 = \left( {4 + 12y + 9{y^2}} \right) - \left( {4{x^2} - 12xy + 9{y^2}} \right) - 12xy\\
 = 4 + 12y + 9{y^2} - 4{x^2} + 12xy - 9{y^2} - 12xy\\
 = 4 + 12y + \left( {9{y^2} - 9{y^2}} \right) + \left( {12xy - 12xy} \right) - 4{x^2}\\
 = 4 + 12y - 4{x^2}\\
d,\\
{\left( {x + 1} \right)^3} - {\left( {x - 1} \right)^3} - \left( {{x^3} - 1} \right) - \left( {x - 1} \right)\left( {{x^2} + x + 1} \right)\\
 = \left( {{x^3} + 3.{x^2}.1 + 3.x{{.1}^2} + {1^3}} \right) - \left( {{x^3} - 3.{x^2}.1 + 3.x{{.1}^2} - {1^3}} \right) - \left( {{x^3} - 1} \right) - \left( {x - 1} \right)\left( {{x^2} + x.1 + {1^2}} \right)\\
 = \left( {{x^3} + 3{x^2} + 3x + 1} \right) - \left( {{x^3} - 3{x^2} + 3x - 1} \right) - \left( {{x^3} - 1} \right) - \left( {{x^3} - {1^3}} \right)\\
 = {x^3} + 3{x^2} + 3x + 1 - {x^3} + 3{x^2} - 3x + 1 - {x^3} + 1 - {x^3} + 1\\
 = \left( {{x^3} - {x^3} - {x^3} - {x^3}} \right) + \left( {3{x^2} + 3{x^2}} \right) + \left( {3x - 3x} \right) + \left( {1 + 1 + 1 + 1} \right)\\
 =  - 2{x^3} + 6{x^2} + 4
\end{array}\)