Question

Rút gọn biểu thức:

\(\dfrac{1}{4}\left( {2{x^2} + y} \right)\left( {x - 2{y^2}} \right) + \dfrac{1}{4}\left( {2{x^2} - y} \right)\left( {x + 2{y^2}} \right)\) 

Answer 1

\(\begin{array}{l}\dfrac{1}{4}\left( {2{x^2} + y} \right)\left( {x - 2{y^2}} \right) + \dfrac{1}{4}\left( {2{x^2} - y} \right)\left( {x + 2{y^2}} \right)\\ = \left( {\dfrac{1}{2}{x^2} + \dfrac{1}{4}y} \right).\left( {x - 2{y^2}} \right) + \left( {\dfrac{1}{2}{x^2} - \dfrac{1}{4}y} \right).\left( {x + 2{y^2}} \right)\\ = \dfrac{1}{2}{x^2}.x - \dfrac{1}{2}{x^2}.2{y^2} + \dfrac{1}{4}y.x - \dfrac{1}{4}y.2{y^2} + \dfrac{1}{2}{x^2}.x + \dfrac{1}{2}{x^2}.2{y^2} - \dfrac{1}{4}y.x - \dfrac{1}{4}y.2{y^2}\\ = \dfrac{1}{2}{x^3} - {x^2}{y^2} + \dfrac{1}{4}xy - \dfrac{1}{2}{y^3} + \dfrac{1}{2}{x^3} + {x^2}{y^2} - \dfrac{1}{4}xy - \dfrac{1}{2}{y^3}\\ = \left( {\dfrac{1}{2}{x^3} + \dfrac{1}{2}{x^3}} \right) + \left( { - \dfrac{1}{2}{y^3} - \dfrac{1}{2}{y^3}} \right) + \left( { - {x^2}{y^2} + {x^2}{y^2}} \right) + \left( {\dfrac{1}{4}xy - \dfrac{1}{4}xy} \right)\\ = {x^3} - {y^3}\end{array}\)