Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
1,\\
\left( {{x^2} + 2xy + {y^2}} \right):\left( {x + y} \right)\\
 = {\left( {x + y} \right)^2}:\left( {x + y} \right)\\
 = \left( {x + y} \right)\\
2,\\
\left( {{x^2} - 4xy + 4{y^2}} \right):\left( {2y - x} \right)\\
 = \left[ {{{\left( {2y} \right)}^2} - 2.2y.x + {x^2}} \right]:\left( {2y - x} \right)\\
 = {\left( {2y - x} \right)^2}:\left( {2y - x} \right)\\
 = 2y - x\\
3,\\
\left( {{x^3} - 8{y^3}} \right):\left( {{x^2} + 2xy + 4{y^2}} \right)\\
 = \left[ {{x^3} - {{\left( {2y} \right)}^3}} \right]:\left( {{x^2} + 2xy + 4{y^2}} \right)\\
 = \left[ {\left( {x - 2y} \right).\left( {{x^2} + x.2y + {{\left( {2y} \right)}^2}} \right)} \right]:\left( {{x^2} + 2xy + 4{y^2}} \right)\\
 = \left[ {\left( {x - 2y} \right).\left( {{x^2} + 2xy + 4{y^2}} \right)} \right]:\left( {{x^2} + 2xy + 4{y^2}} \right)\\
 = x - 2y\\
4,\\
\left( {\frac{1}{8}{x^3} + 27{y^3}} \right):\left( {\frac{1}{2}x + 3y} \right)\\
 = \left[ {{{\left( {\frac{1}{2}x} \right)}^3} + {{\left( {3y} \right)}^3}} \right]:\left( {\frac{1}{2}x + 3y} \right)\\
 = \left[ {\left( {\frac{1}{2}x + 3y} \right).\left[ {{{\left( {\frac{1}{2}x} \right)}^2} - \frac{1}{2}x.3y + {{\left( {3y} \right)}^2}} \right]} \right]:\left( {\frac{1}{2}x + 3y} \right)\\
 = \left[ {\left( {\frac{1}{2}x + 3y} \right).\left( {\frac{1}{4}{x^2} - \frac{3}{2}xy + 9{y^2}} \right)} \right]:\left( {\frac{1}{2}x + 3y} \right)\\
 = \frac{1}{4}{x^2} - \frac{3}{2}xy + 9{y^2}\\
5,\\
\left( {27{x^3} + 27{x^2}y + 9x{y^2} + {y^3}} \right):\left( {9{x^2} + 6xy + {y^2}} \right)\\
 = \left[ {{{\left( {3x} \right)}^3} + 3.{{\left( {3x} \right)}^2}.y + 3.3x.{y^2} + {y^3}} \right]:\left[ {{{\left( {3x} \right)}^2} + 2.3x.y + {y^2}} \right]\\
 = {\left( {3x + y} \right)^3}:{\left( {3x + y} \right)^2}\\
 = 3x + y\\
8,\\
\left( {{x^2} + {y^2} + 2x + 2xy + 2y - 3} \right):\left( {x + y + 3} \right)\\
 = \left[ {\left( {{x^2} + 2xy + {y^2}} \right) + 2.\left( {x + y} \right) - 3} \right]:\left( {x + y + 3} \right)\\
 = \left[ {{{\left( {x + y} \right)}^2} + 2.\left( {x + y} \right) - 3} \right]:\left( {x + y + 3} \right)\\
 = \left[ {\left( {{{\left( {x + y} \right)}^2} + 3.\left( {x + y} \right)} \right) - \left( {\left( {x + y} \right) + 3} \right)} \right]:\left( {x + y + 3} \right)\\
 = \left[ {\left( {x + y} \right)\left( {x + y + 3} \right) - \left( {x + y + 3} \right)} \right]:\left( {x + y + 3} \right)\\
 = \left[ {\left( {x + y + 3} \right)\left( {x + y - 1} \right)} \right]:\left( {x + y + 3} \right)\\
 = x + y - 1
\end{array}\)

Em xem lại đề câu 6,7 nhé, hình như nhầm lẫn dấu + và dấu - rồi