1.如果(x-1)与(y-1)互为倒数,求1/x+1/y的值.
【解】(x-1)(y-1)=1
所以xy-x-y=0;x+y=xy
所以1/x+1/y=(x+y)/(xy)=1
2.设总路程为s;
有如下关系:
s/3a + 2s/6b = 2s/3c
所以a,b,c 的关系为:1/a + 1/b = 2/c
(2)c = 3时,1/a + 1/b = 2/3
1/b - 1/a = 1/6
由上两式解得:a = 4
b = 12/5
1.如果(x-1)与(y-1)互为倒数,求1/x+1/y的值.
【解】(x-1)(y-1)=1
所以xy-x-y=0;x+y=xy
所以1/x+1/y=(x+y)/(xy)=1
2.设总路程为s;
有如下关系:
s/3a + 2s/6b = 2s/3c
所以a,b,c 的关系为:1/a + 1/b = 2/c
(2)c = 3时,1/a + 1/b = 2/3
1/b - 1/a = 1/6
由上两式解得:a = 4
b = 12/5