[(x+1/4y)^2+(x-1/4y)^2](2x^2-1/8y^2) 初一乘法公式的方法解

(x+y)(x-y)+4(y-1) x²/xy-x/y 3x/4x+y - x-2y/4x+y (1-y/y+x)÷x/y²-x² 把4（x-y+1)+y(y-2x)因式分解 y(y+2)+4x(x-y-1)因式分解是多少 {3(x+y)-4(x-y)=4 {x+y/2 + x-y/6=1 把(x+y),(x-y)个作为一个整体合并同类项1,7(x+y)-2(x+y)-4(x+y)+2(x+y)2,(x-y)²＋2(x-y)－(x-y)－2(x-y)² 计算:(4xy(x-y)^-1+(x(y^-1)-(x^-1)y)/((x^-1)+(y^-1)))*(x+y)^-2 x+y+2(-x-y+1)=3(1-y-x)-4-(y+x-1),x+y=? 因式分解：-1/2*x*x*x*y+1/4*x*x*y*y-1/8*x*x*x*y*y*y 计算:1/(x-y)-1/(x+y)-2y/(x^2+y^2)-4y^3/(x^4+y^4)-8x^7/(x^8+y^8) 已知:【(x²+y²)-(x-y)²+2y(x-y)】/4y=1,求4x/4x²-y²-1/2x+y 先化简 再求值 (2x-y)(y+x)-(x-2y)(2y+x)-(-3y+x)^2其(√x+1)+y^2+4=-4y 化简[(3x+4y)^2-(2x+y)(2x-y)+(-x+y)(5x-y)]除以-2y,其中x=-1,y=1 已知x-y=1,求[(x+2y)^2+(2x+y)(x-4y)-3(x+y)(x-y)]除以y的值大神们帮帮忙 已知x-y=1,求[(x+2y)²+(2x+y)(x-4y)-3(x+y)(x-y)]÷y的值 1/(x-y)+1/(x+y)+2x/(x^2+y^2)+4x^3/(x^4+y^4)化简 化简：[(2x-y)平方+（2x+y)(2x-y)-4x(x-y+1)]÷2x （x+y)²-4(x+y-1)