Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
4,\\
a,\\
{x^2} - 8x + 16 = {x^2} - 2.x.4 + {4^2} = {\left( {x - 4} \right)^2}\\
b,\\
9{x^2} + 6x + 1 = {\left( {3x} \right)^2} + 2.3x.1 + {1^2} = {\left( {3x + 1} \right)^2}\\
c,\\
{x^2} - 4xy + 4{y^2} = {x^2} - 2.x.2y + {\left( {2y} \right)^2} = {\left( {x - 2y} \right)^2}\\
d,\\
10xy - 25{y^2} - {x^2} = - \left( {{x^2} - 10xy + 25{y^2}} \right) = - \left( {{x^2} - 2.x.5y + {{\left( {5y} \right)}^2}} \right) = - {\left( {x - 5y} \right)^2}\\
5,\\
a,\\
8{x^3} - 27 = {\left( {2x} \right)^3} - {3^3} = \left( {2x - 3} \right)\left[ {{{\left( {2x} \right)}^2} + 2x.3 + {3^2}} \right] = \left( {2x - 3} \right)\left( {4{x^2} + 6x + 9} \right)\\
b,\\
125{y^3} + 1 = {\left( {5y} \right)^3} + {1^3} = \left( {5y + 1} \right)\left[ {{{\left( {5y} \right)}^2} - 5y.1 + {1^2}} \right] = \left( {5y + 1} \right)\left( {25{y^2} - 5y + 1} \right)\\
c,\\
64{x^3} - 27{y^3} = {\left( {4x} \right)^3} - {\left( {3y} \right)^3} = \left( {4x - 3y} \right)\left[ {{{\left( {4x} \right)}^2} + 4x.3y + {{\left( {3y} \right)}^2}} \right] = \left( {4x - 3y} \right)\left( {16{x^2} + 12xy + 9{y^2}} \right)\\
d,\\
27{x^3} + \frac{{{y^3}}}{8} = {\left( {3x} \right)^3} + {\left( {\frac{y}{2}} \right)^3} = \left( {3x + \frac{y}{2}} \right).\left[ {{{\left( {3x} \right)}^2} - 3x.\frac{y}{2} + {{\left( {\frac{y}{2}} \right)}^2}} \right] = \left( {3x + \frac{y}{2}} \right)\left( {9{x^2} - \frac{3}{2}xy + \frac{{{y^2}}}{4}} \right)\\
e,\\
{\left( {x - 2} \right)^3} + 64 = {\left( {x - 2} \right)^3} + {4^3} = \left[ {\left( {x - 2} \right) + 4} \right]\left[ {{{\left( {x - 2} \right)}^2} - \left( {x - 2} \right).2 + {2^2}} \right] = \left( {x + 2} \right)\left( {{x^2} - 4x + 4 - 2x + 4 + 4} \right) = \left( {x + 2} \right)\left( {{x^2} - 6x + 12} \right)\\
f,\\
125 - {\left( {x + 1} \right)^3} = {5^3} - {\left( {x + 1} \right)^5} = \left[ {5 - \left( {x + 1} \right)} \right].\left[ {{5^2} + 5.\left( {x + 1} \right) + {{\left( {x + 1} \right)}^2}} \right] = \left( {4 - x} \right)\left( {25 + 5x + 5 + {x^2} + 2x + 1} \right) = \left( {4 - x} \right)\left( {{x^2} + 7x + 31} \right)\\
6,\\
a,\\
{x^3} + 6{x^2} + 12x + 8 = {x^3} + 3.{x^2}.2 + 3.x{.2^2} + {2^3} = {\left( {x + 2} \right)^3}\\
b,\\
{x^3} - 3{x^2} + 3x - 1 = {\left( {x - 1} \right)^3}\\
c,\\
1 - 9x + 27{x^2} - 27{x^3} = {1^3} - {3.1^2}.3x + 3.1.{\left( {3x} \right)^2} - {\left( {3x} \right)^3} = {\left( {1 - 3x} \right)^3}
\end{array}\)
