Kết quả thương của phép chia (3xy2−2x2y+x3):(−12x)\left( {3x{y^2} - 2{x^2}y + {x^3}} \right):\left( { - \frac{1}{2}x} \right)(3xy2−2x2y+x3):(−21x) là:
Ta có:
(3xy2−2x2y+x3):(−12x)=3xy2:(−12x)−2x2y:(−12x)+x3:(−12x)=−6y2+4xy−2x2(3xy2−2x2y+x3):(−12x)=3xy2:(−12x)−2x2y:(−12x)+x3:(−12x)=−6y2+4xy−2x2\begin{array}{l}\left( {3x{y^2} - 2{x^2}y + {x^3}} \right):\left( { - \frac{1}{2}x} \right)\\ = 3x{y^2}:\left( { - \frac{1}{2}x} \right) - 2{x^2}y:\left( { - \frac{1}{2}x} \right) + {x^3}:\left( { - \frac{1}{2}x} \right)\\ = - 6{y^2} + 4xy - 2{x^2}\end{array}(3xy2−2x2y+x3):(−21x)=3xy2:(−21x)−2x2y:(−21x)+x3:(−21x)=−6y2+4xy−2x2(3xy2−2x2y+x3):(−21x)=3xy2:(−21x)−2x2y:(−21x)+x3:(−21x)=−6y2+4xy−2x2
