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Phân tích đa thức thành nhân tử:
a) \({x^3} – 2{x^2} + x – x{y^2}\) ;
b) \({x^2} – 7x + 12\) ;
c) \({x^2} – x – 6\) ;
d) \(2{x^2} + x – 6\) ;
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e) \({x^3} – 2x – 4\) .
\(\eqalign{ & a)\,\,{x^3} – 2{x^2} + x – x{y^2} = x\left( {{x^2} – 2x + 1 – {y^2}} \right) \cr & \,\,\,\,\, = x\left[ {{{\left( {x – 1} \right)}^2} – {y^2}} \right] = x\left( {x – 1 – y} \right)\left( {x – 1 + y} \right) \cr & b)\,\,{x^2} – 7x + 12 = {x^2} – 4x – 3x + 12 \cr & \,\,\,\,\, = x\left( {x – 4} \right) – 3\left( {x – 4} \right) = \left( {x – 4} \right)\left( {x – 3} \right) \cr & c)\,\,\,{x^2} – x – 6 = {x^2} – 3x + 2x – 6 \cr & \,\,\,\,\, = x\left( {x – 3} \right) + 2\left( {x – 3} \right) = \left( {x – 3} \right)\left( {x + 2} \right) \cr & d)\,\,2{x^2} + x – 6 = 2{x^2} + 4x – 3x – 6 \cr & \,\,\,\,\, = 2x\left( {x + 2} \right) – 3\left( {x + 2} \right) = \left( {x + 2} \right)\left( {2x – 3} \right) \cr & e)\,\,{x^3} – 2x – 4 = {x^3} – 2x – 8 + 4 \cr & \,\,\,\,\, = \left( {{x^3} – 8} \right) – \left( {2x – 4} \right) = \left( {{x^3} – {2^3}} \right) – 2\left( {x – 2} \right) \cr & \,\,\,\,\, = \left( {x – 2} \right)\left( {{x^2} + 2x + 4} \right) – 2\left( {x – 2} \right) \cr & \,\,\,\,\, = \left( {x – 2} \right)\left( {{x^2} + 2x + 4 – 2} \right) \cr & \,\,\,\,\, = \left( {x – 2} \right)\left( {{x^2} + 2x + 2} \right) \cr} \)