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
\(\eqalign{ & a)\,\,6x + 3 - \left( {2x - 5} \right)\left( {2x + 1} \right) \cr & \,\,\,\,\,\, = 3\left( {2x + 1} \right) - \left( {2x - 5} \right)\left( {2x + 1} \right) \cr & \,\,\,\,\,\, = \left( {2x + 1} \right)\left( {3 - 2x + 5} \right) \cr & \,\,\,\,\,\, = \left( {2x + 1} \right)\left( {8 - 2x} \right) \cr & \,\,\,\,\,\, = 2\left( {2x + 1} \right)\left( {4 - x} \right) \cr & b)\,\,a{b^3}{c^2} - {a^2}{b^2}{c^2} + a{b^2}{c^3} - {a^2}b{c^3} \cr & \,\,\,\,\,\, = ab{c^2}\left( {{b^2} - ab + bc - ac} \right) \cr & \,\,\,\,\,\, = ab{c^2}\left[ {\left( {{b^2} - ab} \right) + \left( {bc - ac} \right)} \right] \cr & \,\,\,\,\,\, = ab{c^2}\left[ {b\left( {b - a} \right) + c\left( {b - a} \right)} \right] \cr & \,\,\,\,\,\, = ab{c^2}\left( {b - a} \right)\left( {b + c} \right) \cr & c)\,\,a{x^2} + c{x^2} - ay + a{y^2} - cy + c{y^2} \cr & \,\,\,\,\, = \left( {a{x^2} - ay} \right) + \left( {c{x^2} - cy} \right) + \left( {a{y^2} + c{y^2}} \right) \cr & \,\,\,\,\, = a\left( {{x^2} - y} \right) + c\left( {{x^2} - y} \right) + {y^2}\left( {a + c} \right) \cr & \,\,\,\,\, = \left( {{x^2} - y} \right)\left( {a + c} \right) + {y^2}\left( {a + c} \right) \cr & \,\,\,\,\, = \left( {a + c} \right)\left( {{x^2} - y + {y^2}} \right) \cr & d)\,\,5x{y^3} - 2xyz - 15{y^2} + 6z \cr & \,\,\,\,\,\, = \left( {5x{y^3} - 15{y^2}} \right) - \left( {2xyz - 6z} \right) \cr & \,\,\,\,\,\, = 5{y^2}\left( {xy - 3} \right) - 2z\left( {xy - 3} \right) \cr & \,\,\,\,\,\, = \left( {3x - y} \right)\left( {5{y^2} - 2z} \right) \cr} \)
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