数学分式题帮我解下，太难了，解不出来！！！！！！！！！！！！

显示全部楼层 · 2013-7-19 11:07:17

已知x2+y2-8x-10y+41=0那么，已知x2-8x+16+y2-10y+25=0即：(x-4)2+(y-5)2=0则x-4=0,x=4; y-5=0,y=5所以，y/x2-xy +x/y2-xy=5/(42-4×5)+4/(52-4×5)=-5/4+4/5=-9/20...

千问 · 2013-7-19 11:07:17

答：x2+y2-8x-10y+41=0x2-8x+16+y2-10y+25=0(x-4)2+(y-5)2=0所以：x-4=0，y-5=0所以：x=4，y=5所以：y/(x2-xy)+x/(y2-xy)=y2/[xy(x-y)]-x&#17...

千问 · 2013-7-19 11:07:17

x2+y2-8x-10y+41=0(x2-8x+16)+y2-10y+25=0(x-4)2+(y-5)2=0x-4=0, y-5=0x=4,y=5y/x2-xy+x/y2-xy=5/(16-20)+4/(25-20)=-5/4+4/5...

千问 · 2013-7-19 11:07:17

已知x2+y2-8x-10y+41=0,求y/(x2-xy ) +x(/y2-xy)的值解：x2+y2-8x-10y+41=(x-4)2+(y-5)2=0，由于(x-4)2≧0，(y-5)2≧0；现(x-4)2+(y-5)2=0，...

千问 · 2013-7-19 11:07:17

x2+y2-8x-10y+41=x2-8x+16+y2-10y+25=(x-4)2+(y-5)2=0解得x=4，y=5带入得原式=-9/20...