Advertisements (Quảng cáo)
Phân tích các đa thức sau thành nhân tử
a) \({x^2} – 3x + 2\) ;
b) \({x^2} – x – 6\) ;
c) \({x^2} + 7x + 12\) ;
d) \(3{x^2} + x – 2\) ;
e) \({x^4} + {x^2} + 1\) ;
f) \({x^2} + 2xy – 15{y^2}\) .
\(\eqalign{ & a)\,\,{x^2} – 3x + 2 \cr & \,\,\,\,\, = {x^2} – x – 2x + 2 \cr & \,\,\,\, = \left( {{x^2} – x} \right) – \left( {2x – 2} \right) \cr & \,\,\,\, = x\left( {x – 1} \right) – 2\left( {x – 1} \right) \cr & \,\,\,\, = \left( {x – 1} \right)\left( {x – 2} \right) \cr & b)\,\,{x^2} – x – 6 \cr & \,\,\,\,\, = {x^2} – 3x + 2x – 6 \cr & \,\,\,\,\, = \left( {{x^2} – 3x} \right) + \left( {2x – 6} \right) \cr & \,\,\,\,\, = x\left( {x – 3} \right) + 2\left( {x – 3} \right) \cr & \,\,\,\,\, = \left( {x – 3} \right)\left( {x + 2} \right) \cr & c)\,\,{x^2} + 7x + 12 \cr & \,\,\,\,\, = {x^2} + 3x + 4x + 12 \cr & \,\,\,\,\, = \left( {{x^2} + 3x} \right) + \left( {4x + 12} \right) \cr & \,\,\,\,\, = x\left( {x + 3} \right) + 4\left( {x + 3} \right) \cr & \,\,\,\,\, = \left( {x + 3} \right)\left( {x + 4} \right) \cr & d)\,\,3{x^2} + x – 2 \cr & \,\,\,\,\, = 3{x^2} + 3x – 2x – 2 \cr & \,\,\,\,\, = \left( {3{x^2} + 3x} \right) – \left( {2x + 2} \right) \cr & \,\,\,\,\, = 3x\left( {x + 2} \right) – 2\left( {x + 1} \right) \cr & \,\,\,\,\, = \left( {x + 1} \right)\left( {3x – 2} \right) \cr & e)\,\,{x^4} + {x^2} + 1 \cr & \,\,\,\,\, = {x^4} + {x^2} + 1 + {x^2} – {x^2} \cr & \,\,\,\,\, = \left( {{x^4} + 2{x^2} + 1} \right) – {x^2} \cr & \,\,\,\,\, = {\left( {{x^2} + 1} \right)^2} – {x^2} \cr & \,\,\,\,\, = \left( {{x^2} + 1 – x} \right)\left( {{x^2} + 1 + x} \right) \cr & f)\,\,{x^2} + 2xy – 15{y^2} \cr & \,\,\,\,\, = {x^2} + 2xy + {y^2} – 16{y^2} \cr & \,\,\,\,\, = \left( {{x^2} + 2xy + {y^2}} \right) – 16{y^2} \cr & \,\,\,\,\, = {\left( {x + 2y} \right)^2} – {\left( {4y} \right)^2} \cr & \,\,\,\,\, = \left( {x + 2y – 4y} \right)\left( {x + 2y + 4y} \right) \cr & \,\,\,\,\, = \left( {x – 2y} \right)\left( {x + 6y} \right) \cr} \)