y2+(x2+2xy-3y2)-(2x2-xy-2y2) 这个代数式的正确写法及解法

求证:(x2-xy+y2)3+(x2+xy+y2)3能被2x2+2y2整除 若X2+xy-2y2=0,则（x2+3xy+y2)/x2+y2=? 4x2-3y2+2xy-4x2+5x2 (x2+2xy+y2)( x2-xy+y2)2 化简 (x2-2xy+y2)-(x2+2xy+y2)=什么 ( x2+xy+y2) ( x2+xy+2y2)-12y4 x2-2xy+y2-9 2x2-xy+3y2因式分解 x2 y2+3xy+2 因式分解 x2-3y2=2xy过程 2X2-3XY-Y2分解因式! 4x2+3xy-2y2因式分解 -（x2-2xy-y2)+(5x2-2xy-3y2)-（x2-2xy-y2)+(5x2-2xy-3y2)化简 2(x2-xy)-3(2x2-3xy)-2[x2-(2x2-xy+y2)] 化简. 已知x2+xy=3,xy+y2=-2,则2x2-xy-3y2= 已知x2-3xy=-5,xy+y2=3,求x2-2xy+y2. 如果x2+y2=5,那么代数式(3x2-2xy-y2)-(x2-2xy-3y2)的值为 (x2+y2)(x2-xy+y2)-2x2y2因式分解