Đáp án:

\(\begin{array}{l}
1,\\
{x^2} + 8x + 15\\
2,\\
{x^2} - 3x - 4\\
3,\\
2{x^2} + 9x - 18\\
4,\\
{x^3} - 3{x^2} + 2x - 6\\
5,\\
6{x^2} - 23x + 20\\
6,\\
{x^2} - \dfrac{1}{4}{y^2}\\
7,\\
{x^3} - {x^2} - 3x + 2\\
8,\\
24{x^2} + 4x - 8
\end{array}\)

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
1,\\
\left( {x + 3} \right)\left( {x + 5} \right) = x.\left( {x + 5} \right) + 3.\left( {x + 5} \right)\\
 = {x^2} + 5x + 3x + 15 = {x^2} + 8x + 15\\
2,\\
\left( {x + 1} \right)\left( {x - 4} \right) = x.\left( {x - 4} \right) + 1.\left( {x - 4} \right)\\
 = {x^2} - 4x + x - 4 = {x^2} - 3x - 4\\
3,\\
\left( {2x - 3} \right)\left( {x + 6} \right) = 2x.\left( {x + 6} \right) - 3.\left( {x + 6} \right)\\
 = \left( {2{x^2} + 12x} \right) - \left( {3x + 18} \right)\\
 = 2{x^2} + 12x - 3x - 18\\
 = 2{x^2} + 9x - 18\\
4,\\
\left( {{x^2} + 2} \right)\left( {x - 3} \right) = {x^2}.\left( {x - 3} \right) + 2.\left( {x - 3} \right)\\
 = {x^3} - 3{x^2} + 2x - 6\\
5,\\
\left( {3x - 4} \right)\left( {2x - 5} \right) = 3x.\left( {2x - 5} \right) - 4.\left( {2x - 5} \right)\\
 = \left( {6{x^2} - 15x} \right) - \left( {8x - 20} \right)\\
 = 6{x^2} - 15x - 8x + 20\\
 = 6{x^2} - 23x + 20\\
6,\\
\left( {x - \dfrac{1}{2}y} \right)\left( {x + \dfrac{1}{2}y} \right) = {x^2} - {\left( {\dfrac{1}{2}y} \right)^2} = {x^2} - \dfrac{1}{4}{y^2}\\
7,\\
\left( {{x^2} + x - 1} \right)\left( {x - 2} \right) = \left( {x - 2} \right).\left( {{x^2} + x - 1} \right)\\
 = x.\left( {{x^2} + x - 1} \right) - 2.\left( {{x^2} + x - 1} \right)\\
 = \left( {{x^3} + {x^2} - x} \right) - \left( {2{x^2} + 2x - 2} \right)\\
 = {x^3} + {x^2} - x - 2{x^2} - 2x + 2\\
 = {x^3} - {x^2} - 3x + 2\\
8,\\
4\left( {2x - 1} \right)\left( {3x + 2} \right)\\
 = 4.\left[ {2x\left( {3x + 2} \right) - 1.\left( {3x + 2} \right)} \right]\\
 = 4.\left[ {\left( {6{x^2} + 4x} \right) - \left( {3x + 2} \right)} \right]\\
 = 4.\left( {6{x^2} + 4x - 3x - 2} \right)\\
 = 4.\left( {6{x^2} + x - 2} \right)\\
 = 24{x^2} + 4x - 8
\end{array}\)