当xy≠0时,试说明代数式[(x+y)(x-y)-(x+y)²-2y(x-y)-2xy]/xy的值与x,y的值
2个回答
[(x+y)(x-y)-(x+y)²-2y(x-y)-2xy]/xy
=[x²-y²-x²-2xy-y²-2xy+2y²-2xy]/xy
=-6xy/xy
=-6
相关问题
设|x-2|+(y+¼)²=0,试求代数式x²(x²-xy+y)-x(x³-2x²y+xy-1)的值.
已知x+3xy-y=0,求代数式2x-14xy-2y/x-2xy-y的值
若x-2y=0,求代数式-x²+xy+y²分之2x²-xy-y²的值.
已知x+y=1,xy=-2,试求代数式(xy+5x-2y)-(-3xy+2x-5y)的值.
1.若x+y=1,xy=-2,试求代数式(xy+5x-2y)-(-3xy+2x-5y)的值
当x=2,y=12时,求代数式(x+y)(x-y)+(x-y)2-(x2-3xy)的值.
诚说明不论x,y取何值时,代数式(x³+3x²-5xy²+6y³+1)-(2x³-y³-2xy²-x²y-2)-(4x²
已知x²+y²=2,xy=-½,求代数式(2x²-y²-3xy)-﹙x²-2y²+xy)的值
已知X+Y=-1,xy=-2,求代数式-5(x+y)+(x-y)+(xy+y)的值
x-y+3xy=0 求2x+3xy-2y/x-2xy-y的值