Đề bài
Phân tích đa thức thành nhân tử:
a) \({x^2} - xy + x - y\) ;
b) \({a^3} - {a^2}x - ay + xy\) ;
c) \(3{x^2} + 12x + 12\) ;
d) \(2{a^2} - 2{b^2} - 5a + 5b\) ;
e) \({x^2}y - x{y^2} - 3{x^2} + 3{y^2}\) ;
f) \({x^3} + 5{x^2} - 4x - 20\) .
Lời giải chi tiết
\(\eqalign{ & a)\,\,{x^2} - xy + x - y \cr & \,\,\,\,\, = \left( {{x^2} - xy} \right) + \left( {x - y} \right) \cr & \,\,\,\,\, = x\left( {x - y} \right) + \left( {x - y} \right) \cr & \,\,\,\,\, = \left( {x - y} \right)\left( {x + 1} \right) \cr & b)\,\,{a^3} - {a^2}x - ay + xy \cr & \,\,\,\,\, = \left( {{a^3} - {a^2}x} \right) - \left( {ay - xy} \right) \cr & \,\,\,\,\, = {a^2}\left( {a - x} \right) - y\left( {a - x} \right) \cr & \,\,\,\,\, = \left( {a - x} \right)\left( {{a^2} - y} \right) \cr & c)\,\,3{x^2} + 12x + 12 \cr & \,\,\,\,\, = 3\left( {{x^2} + 4x + 4} \right) \cr & \,\,\,\,\, = 3{\left( {x + 2} \right)^2} \cr & d)\,\,2{a^2} - 2{b^2} - 5a + 5b \cr & \,\,\,\,\, = 2\left( {{a^2} - {b^2}} \right) - \left( {5a - 5b} \right) \cr & \,\,\,\,\, = 2\left( {a - b} \right)\left( {a + b} \right) - 5\left( {a - b} \right) \cr & \,\,\,\,\, = \left( {a - b} \right)\left( {2a + 2b - 5} \right) \cr & e)\,\,{x^2}y - x{y^2} - 3{x^2} + 3{y^2} \cr & \,\,\,\,\, = \left( {{x^2}y - x{y^2}} \right) - \left( {3{x^2} - 3{y^2}} \right) \cr & \,\,\,\,\, = xy\left( {x - y} \right) - 3\left( {{x^2} - {y^2}} \right) \cr & \,\,\,\,\, = xy\left( {x - y} \right) - 3\left( {x - y} \right)\left( {x + y} \right) \cr & \,\,\,\,\, = \left( {x - y} \right)\left[ {xy - 3\left( {x + y} \right)} \right] \cr & \,\,\,\,\, = \left( {x - y} \right)\left( {xy - 3x - 3y} \right) \cr & f)\,\,{x^3} + 5{x^2} - 4x - 20 \cr & \,\,\,\,\, = \left( {{x^3} + 5{x^2}} \right) - \left( {4x + 20} \right) \cr & \,\,\,\,\, = {x^2}\left( {x + 5} \right) - 4\left( {x + 5} \right) \cr & \,\,\,\,\, = \left( {x + 5} \right)\left( {{x^2} - 4} \right) \cr & \,\,\,\,\, = \left( {x + 5} \right)\left( {x - 2} \right)\left( {x + 2} \right) \cr} \)
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