已知:x2-y2=16,x+y=2,求:3x-3y/y2-x2+x2-2xy+y2/x2-y2-x2-x-6/x+2除以(x+y)
问题描述:
已知:x2-y2=16,x+y=2,求:3x-3y/y2-x2+x2-2xy+y2/x2-y2-x2-x-6/x+2除以(x+y)
答
已知:x2-y2=16,x+y=2,
求:(3x-3y)/(y2-x2)+(x2-2xy+y2)/(x2-y2)-(x2-x-6)/(x+2)
除以(x+y)
由x2-y2=16,x+y=2,可得:x-y=8;
则::
(3x-3y)/(y2-x2)+(x2-2xy+y2)/(x2-y2)-(x2-x-6)/(x+2)/(x+y)
=24/16+64/16-(x-3)/2
=1.5+4-x/2+1.5
=7-x/2 .