Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
g,\\
{a^3} - {a^2} - a + 1\\
= \left( {{a^3} - {a^2}} \right) - \left( {a - 1} \right)\\
= {a^2}\left( {a - 1} \right) - \left( {a - 1} \right)\\
= \left( {a - 1} \right)\left( {{a^2} - 1} \right)\\
= \left( {a - 1} \right)\left( {a - 1} \right)\left( {a + 1} \right)\\
= {\left( {a - 1} \right)^2}.\left( {a + 1} \right)\\
h,\\
{m^2} + am + ay - {y^2}\\
= \left( {{m^2} - {y^2}} \right) + \left( {am + ay} \right)\\
= \left( {m - y} \right)\left( {m + y} \right) + a\left( {m + y} \right)\\
= \left( {m + y} \right).\left( {m - y + a} \right)\\
i,\\
3xy + {y^2} - 3x - 1\\
= \left( {3xy - 3x} \right) + \left( {{y^2} - 1} \right)\\
= 3x\left( {y - 1} \right) + \left( {y - 1} \right)\left( {y + 1} \right)\\
= \left( {y - 1} \right)\left( {3x + y + 1} \right)\\
k,\\
{x^3} - x{y^2} + {x^2}y - {y^3}\\
= \left( {{x^3} + {x^2}y} \right) - \left( {x{y^2} + {y^3}} \right)\\
= {x^2}\left( {x + y} \right) - {y^2}\left( {x + y} \right)\\
= \left( {x + y} \right).\left( {{x^2} - {y^2}} \right)\\
= {\left( {x + y} \right)^2}.\left( {x - y} \right)\\
l,\\
{a^3} - ma - mb + {b^3}\\
= \left( {{a^3} + {b^3}} \right) - \left( {ma + mb} \right)\\
= \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right) - m\left( {a + b} \right)\\
= \left( {a + b} \right).\left( {{a^2} - ab + {b^2} - m} \right)
\end{array}\)
