[(x^2-2xy)+x^2]/(y^2+2xy)-(x^2+2xy)
1个回答
原式=(2x²-2xy)/(y²+2xy-x²-2xy)
=2x(x-y)/(y+x)(y-x)
=-2x/(x+y)
相关问题
x2+2xy+y2x2y+xy2−x2−2xy+y2x2y−xy2.
3x2y−[2xy2−2(xy−32x2y)+xy]+3xy2.
5xy^2-(3xy^2-(4xy^2-2x^2y))+2x^2y-xy^2
2(x²-xy)-3(2x²-3xy)-2[x²-(2x²-xy+y²
-2x^2y(-2xy^2)^2+(2xy)^3X(xy^2)
2xy^2+x^2y-xy=xy( )
3xy(x^2-xy^2+xy)-xy^2(2x^2-xy+2x)
如果x2+xy=2,xy+y2=-1,则x2-y2=______,x2+2xy+y2=______.
3x^2 y+{xy-[3x^2 y-(4xy^2 +1/2xy)]-(4x^2 y+3/2xy)}
计算:(x|x^2-xy)^﹣3×(﹣x^2-2xy+y^2)﹣2÷(x^2|xy-y^2)^2=