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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) \(2{a^2} – 2{b^2} – 5a + 5b\) ;
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e) \({x^2}y – x{y^2} – 3{x^2} + 3{y^2}\) ;
f) \({x^3} + 5{x^2} – 4x – 20\) .
\(\eqalign{ & a)\,\,{x^2} – xy + x – y \cr & \,\,\,\,\, = \left( {{x^2} – xy} \right) + \left( {x – y} \right) \cr & \,\,\,\,\, = x\left( {x – y} \right) + \left( {x – y} \right) \cr & \,\,\,\,\, = \left( {x – y} \right)\left( {x + 1} \right) \cr & b)\,\,{a^3} – {a^2}x – ay + xy \cr & \,\,\,\,\, = \left( {{a^3} – {a^2}x} \right) – \left( {ay – xy} \right) \cr & \,\,\,\,\, = {a^2}\left( {a – x} \right) – y\left( {a – x} \right) \cr & \,\,\,\,\, = \left( {a – x} \right)\left( {{a^2} – y} \right) \cr & c)\,\,3{x^2} + 12x + 12 \cr & \,\,\,\,\, = 3\left( {{x^2} + 4x + 4} \right) \cr & \,\,\,\,\, = 3{\left( {x + 2} \right)^2} \cr & d)\,\,2{a^2} – 2{b^2} – 5a + 5b \cr & \,\,\,\,\, = 2\left( {{a^2} – {b^2}} \right) – \left( {5a – 5b} \right) \cr & \,\,\,\,\, = 2\left( {a – b} \right)\left( {a + b} \right) – 5\left( {a – b} \right) \cr & \,\,\,\,\, = \left( {a – b} \right)\left( {2a + 2b – 5} \right) \cr & e)\,\,{x^2}y – x{y^2} – 3{x^2} + 3{y^2} \cr & \,\,\,\,\, = \left( {{x^2}y – x{y^2}} \right) – \left( {3{x^2} – 3{y^2}} \right) \cr & \,\,\,\,\, = xy\left( {x – y} \right) – 3\left( {{x^2} – {y^2}} \right) \cr & \,\,\,\,\, = xy\left( {x – y} \right) – 3\left( {x – y} \right)\left( {x + y} \right) \cr & \,\,\,\,\, = \left( {x – y} \right)\left[ {xy – 3\left( {x + y} \right)} \right] \cr & \,\,\,\,\, = \left( {x – y} \right)\left( {xy – 3x – 3y} \right) \cr & f)\,\,{x^3} + 5{x^2} – 4x – 20 \cr & \,\,\,\,\, = \left( {{x^3} + 5{x^2}} \right) – \left( {4x + 20} \right) \cr & \,\,\,\,\, = {x^2}\left( {x + 5} \right) – 4\left( {x + 5} \right) \cr & \,\,\,\,\, = \left( {x + 5} \right)\left( {{x^2} – 4} \right) \cr & \,\,\,\,\, = \left( {x + 5} \right)\left( {x – 2} \right)\left( {x + 2} \right) \cr} \)