Đề bài
Phân tích các đa thức thành nhân tử:
a) \(5{x^2} - 10x + 5\) ;
b) \( - 3{x^2} + 6xy - 3{y^2}\) ;
c) \({a^2} - {b^2} + 2a - 2b\) ;
d) \({x^3} - 6{x^2}y + 9x{y^2}\) ;
e) \({x^2} - 4xy + 4{y^2} - 9\) .
Lời giải chi tiết
\(\eqalign{ & a)\,\,5{x^2} - 10x + 5 \cr & \,\,\,\,\,\, = 5\left( {{x^2} - 2x + 1} \right) \cr & \,\,\,\,\,\, = 5{\left( {x - 1} \right)^2} \cr & b)\,\, - 3{x^2} + 6xy - 3{y^2} \cr & \,\,\,\,\, = - 3\left( {{x^2} - 2xy + {y^2}} \right) \cr & \,\,\,\,\, = - 3{\left( {x - y} \right)^2} \cr & c)\,\,{a^2} - {b^2} + 2a - 2b = \left( {{a^2} - {b^2}} \right) + \left( {2a - 2b} \right) \cr & \,\,\,\, = \left( {a - b} \right)\left( {a + b} \right) + 2\left( {a - b} \right) \cr & \,\,\,\, = \left( {a - b} \right)\left( {a + b + 2} \right) \cr & d)\,\,{x^3} - 6{x^2}y + 9x{y^2} \cr & \,\,\,\,\, = x\left( {{x^2} - 6xy + 9{y^2}} \right) \cr & \,\,\,\, = x\left[ {{x^2} - 2.x.3y + {{\left( {3y} \right)}^2}} \right] \cr & \,\,\,\, = x{\left( {x - 3y} \right)^2} \cr & e)\,\,{x^2} - 4xy + {4y^2} - 9 \cr & \,\,\,\, = \left( {{x^2} - 4xy + {4y^2}} \right) - 9 \cr & \,\,\,\, = \left[ {{x^2} - 2.x.2y + {{\left( {2y} \right)}^2}} \right] - 9 \cr & \,\,\,\, = {\left( {x - 2y} \right)^2} - {3^2} \cr & \,\,\,\, = \left( {x - 2y - 3} \right)\left( {x - 2y + 3} \right) \cr} \)
soanvan.me