若2x²+2xy+y²-4x+z-2√z-3 +2=0,求xy+yz+xz的值标注：√为根号,根号下是 x-3

若x²+xy-2y²=0,则x²+3xy+y²/x²+y² x²-2xy+y²-9 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 x²-4y²/x²+2xy+y²÷x+2y/2x²+2xy 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 因式分解 x²-XY-2y²-X-Y x²+xy-2y²-x+7y-6 （xy-x²）÷[（x²-2xy+y²）÷xy]*[(x-y)÷x²] 6x²-xy-15y²=0 求x²-2xy+4y²/2x²-3xy-3y²告诉下 6x²-xy-15y²=0 求2x²-3xy-3y²分之x²-2xy+4y² 谢谢高手解答x²-xy=2,-xy+y²=5求x²-2xy+ y²,x²- y² 若(x+y-5)²+(xy-6)²=0,则（x+y)²-2xy的值为? 已知：x²+xy=3,xy+y²=-2 求：2x²-xy-3y² 因式分解2x(x-y)^4 - x²(x-y)² + xy(y-x)²RT (x+2y)²-x²-2xy