Đề bài
Phân tích đa thức sau thành nhân tử:
a] \[6x + 3 - [2x - 5][2x + 1]\] ;
b] \[a{b^3}{c^2} - {a^2}{b^2}{c^2} + a{b^2}{c^3} - {a^2}b{c^3}\] ;
c] \[a{x^2} + c{x^2} - ay + a{y^2} - cy + c{y^2}\] ;
d] \[5x{y^3} - 2xyz - 15{y^2} + 6z\] .
Lời giải chi tiết
\[\eqalign{ & a]\,\,6x + 3 - \left[ {2x - 5} \right]\left[ {2x + 1} \right] \cr & \,\,\,\,\,\, = 3\left[ {2x + 1} \right] - \left[ {2x - 5} \right]\left[ {2x + 1} \right] \cr & \,\,\,\,\,\, = \left[ {2x + 1} \right]\left[ {3 - 2x + 5} \right] \cr & \,\,\,\,\,\, = \left[ {2x + 1} \right]\left[ {8 - 2x} \right] \cr & \,\,\,\,\,\, = 2\left[ {2x + 1} \right]\left[ {4 - x} \right] \cr & b]\,\,a{b^3}{c^2} - {a^2}{b^2}{c^2} + a{b^2}{c^3} - {a^2}b{c^3} \cr & \,\,\,\,\,\, = ab{c^2}\left[ {{b^2} - ab + bc - ac} \right] \cr & \,\,\,\,\,\, = ab{c^2}\left[ {\left[ {{b^2} - ab} \right] + \left[ {bc - ac} \right]} \right] \cr & \,\,\,\,\,\, = ab{c^2}\left[ {b\left[ {b - a} \right] + c\left[ {b - a} \right]} \right] \cr & \,\,\,\,\,\, = ab{c^2}\left[ {b - a} \right]\left[ {b + c} \right] \cr & c]\,\,a{x^2} + c{x^2} - ay + a{y^2} - cy + c{y^2} \cr & \,\,\,\,\, = \left[ {a{x^2} - ay} \right] + \left[ {c{x^2} - cy} \right] + \left[ {a{y^2} + c{y^2}} \right] \cr & \,\,\,\,\, = a\left[ {{x^2} - y} \right] + c\left[ {{x^2} - y} \right] + {y^2}\left[ {a + c} \right] \cr & \,\,\,\,\, = \left[ {{x^2} - y} \right]\left[ {a + c} \right] + {y^2}\left[ {a + c} \right] \cr & \,\,\,\,\, = \left[ {a + c} \right]\left[ {{x^2} - y + {y^2}} \right] \cr & d]\,\,5x{y^3} - 2xyz - 15{y^2} + 6z \cr & \,\,\,\,\,\, = \left[ {5x{y^3} - 15{y^2}} \right] - \left[ {2xyz - 6z} \right] \cr & \,\,\,\,\,\, = 5{y^2}\left[ {xy - 3} \right] - 2z\left[ {xy - 3} \right] \cr & \,\,\,\,\,\, = \left[ {3x - y} \right]\left[ {5{y^2} - 2z} \right] \cr} \]