Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
{x^2} + 5x - {y^2} - 5y\\
= \left( {{x^2} - {y^2}} \right) + \left( {5x - 5y} \right)\\
= \left( {x - y} \right)\left( {x + y} \right) + 5\left( {x - y} \right)\\
= \left( {x - y} \right)\left( {x + y + 5} \right)\\
b,\\
2{x^2} + 4xy + x + 2y\\
= \left( {2{x^2} + 4xy} \right) + \left( {x + 2y} \right)\\
= 2x\left( {x + 2y} \right) + \left( {x + 2y} \right)\\
= \left( {x + 2y} \right)\left( {2x + 1} \right)\\
c,\\
xy + x + {x^2}y + 1\\
= \left( {xy + {x^2}y} \right) + \left( {x + 1} \right)\\
= xy\left( {1 + x} \right) + \left( {x + 1} \right)\\
= \left( {x + 1} \right)\left( {xy + 1} \right)\\
d,\\
{x^2} - 4 + {\left( {x - 2} \right)^2}\\
= \left( {x - 2} \right)\left( {x + 2} \right) + {\left( {x - 2} \right)^2}\\
= \left( {x - 2} \right).\left[ {\left( {x + 2} \right) + \left( {x - 2} \right)} \right]\\
= \left( {x - 2} \right).2x\\
= 2x\left( {x - 2} \right)\\
e,\\
{x^2} - 4x - 4{y^2} + 4\\
= \left( {{x^2} - 4x + 4} \right) - 4{y^2}\\
= {\left( {x - 2} \right)^2} - {\left( {2y} \right)^2}\\
= \left( {x - 2 - 2y} \right)\left( {x - 2 + 2y} \right)\\
f,\\
45 + {x^3} - 5{x^2} - 9x\\
= \left( {{x^3} - 5{x^2}} \right) - \left( {9x - 45} \right)\\
= {x^2}\left( {x - 5} \right) - 9\left( {x - 5} \right)\\
= \left( {x - 5} \right)\left( {{x^2} - 9} \right)\\
= \left( {x - 5} \right)\left( {x - 3} \right)\left( {x + 3} \right)\\
g,\\
{x^3} - 3{x^2} - 4x + 12\\
= \left( {{x^3} - 3{x^2}} \right) - \left( {4x - 12} \right)\\
= {x^2}\left( {x - 3} \right) - 4\left( {x - 3} \right)\\
= \left( {x - 3} \right)\left( {{x^2} - 4} \right)\\
= \left( {x - 3} \right)\left( {x - 2} \right)\left( {x + 2} \right)\\
h,\\
{x^3} - 3{x^2} + 6x - 18\\
= \left( {{x^3} - 3{x^2}} \right) + \left( {6x - 18} \right)\\
= {x^2}\left( {x - 3} \right) + 6\left( {x - 3} \right)\\
= \left( {x - 3} \right)\left( {{x^2} + 6} \right)\\
i,\\
{x^3} - 4{x^2} - 12x + 27\\
= \left( {{x^3} + 27} \right) - \left( {4{x^2} + 12x} \right)\\
= \left( {x + 3} \right)\left( {{x^2} - 3x + 9} \right) - 4x\left( {x + 3} \right)\\
= \left( {x + 3} \right)\left( {{x^2} - 7x + 9} \right)
\end{array}\)
