1,
[(x-1)/(x+1)+2x/(x²-1)]/[1/(x²-1)]
=[(x-1)/(x+1)+2x/(x²-1)](x²-1)
=[(x-1)/(x+1)+2x/(x²-1)](x+1)(x-1)
=(x-1)²+2x
=x²-2x+1+2x
=x²+1
原式=x²+1,计算结果与x=√2008或x=-√2008的正负无关.
2,
已知A=1/(x-2) ,B=2/(x²-4),C=x/(x+2),
(A-B)/C
=[1/(x-2) -2/(x²-4)]/[x/(x+2)]
={[(x+2)/[(x-2)(x+2) ]-2/[(x+2)(x-2)]}/[x/(x+2)]
=[(x+2-2)/[(x-2)(x+2) ]/[x/(x+2)]
=1/(x-2)
=1/(3-2)
=1
3,
(1)
甲两次购买粮食共需付款:100x+100y=100(x+y)(元);
乙两次共购买粮食为:100/x+100/y(千克);
甲两次购买的平均价格为m元/千克:
m=甲两次购买粮食共需付款数/甲两次购买粮食总质量
=100(x+y)/(100+100)=(x+y)/2(元/千克);
乙两次购买的平均价为n元/千克:
n=乙两次购买共需付款数/乙两次购买粮食总质量
=(100+100)/(100/x+100/y)
=2/(1/x+1/y)
=2xy/(x+y)(元/千克);
(2)比较甲乙两次购买的平均价格即可知道两人的购粮方式哪一种更合算,
m-n
=(x+y)/2-2xy/(x+y)
=[(x+y)²-4xy]/[2(x+y)]
=(x²-2xy+y²-4xy)/[2(x+y)]
=(x-y)²/[2(x+y)]
x>0,y>0,且x≠y
所以(x-y)²/[2(x+y)]>0,m-n>0,即m>n,
所以甲购买的平均价格比乙购买的平均价格高,因此乙的购粮方式更合算.
