x²+xy+y²可以替换成别的代数式吗是一道题目.(x²+xy+y²)(x²-xy+y²)-(x²+y²)²稍微详细点恩,,我们老师不允许我们跳步骤,

x²+xy+y²可以替换成别的代数式吗是一道题目.(x²+xy+y²)(x²-xy+y²)-(x²+y²)²稍微详细点恩,,我们老师不允许我们跳步骤, 0.5x²+xy+0.5y² 因式分解：x²-xy-12y² 4x²y-xy² -9x²-y²+6xy x²-2xy+y²-9 6xy²-9x²y-y²因式分解 若x²+xy-2y²=0,则x²+3xy+y²/x²+y² x²+xy-2y²=0则x²+3xy+y²/x²+y² 因式分解 x²-2xy+y²-z² 计算：4xy²-3x²y-{3x²y-[2xy²-4x²y+2(x²y-2xy²)]急 （x²+y²)²-x²y²=(x²+y²)²-(xy)²=(x²+y²+xy)(x²+y²-xy)=(x+y)²(x-y)²-xy+xy=(x+y)²(x-y)² x²-4y²/x²+2xy+y²÷x+2y/2x²+2xy x²y²-x²y-xy²+x+y-1因式分解 已知x²y²+x²+y²=10xy-16 求x,y 计算x²-x/x²×x/1-x和x²-4y²/x²+2xy+y²÷x+2y/x²+xy x²-y²+6x+9分之x²+y²-9-2xy 3x²-y²=8x²＋xy＋y²=4