1/a+1/b-9/2(a+b)=0,则b/a-a/b=

2014-01-03 00:00

问题补充:
zsm743
zsm743 ·
0 0

1/a+1/b-9/2(a+b)=0 (a+b)/ab=9/2(a+b) 2(a+b)^2=9ab 化简得到: 2a^2-5ab+2b^2=0 即: (2a-b)(a-2b)=0 所以a=2b,或者a=b/2. b/a-a/b =(b^2-a^2)/ab 当a=2b时候,所求式=(b^2-4b^2)/2b^2=-3/2. 当a=b/2时候,所求式=(b^2-b^2/4)/(b^2/2)=3/2.

zsm74
zsm74 ·
0 0

1/a+1/b-9/2(a+b)=0 (a+b)/ab=9/2(a+b) 2(a+b)^2=9ab 化简得到: 2a^2-5ab+2b^2=0 即: (2a-b)(a-2b)=0 所以a=2b,或者a=b/2. b/a-a/b =(b^2-a^2)/ab 当a=2b时候,所求式=(b^2-4b^2)/2b^2=-3/2. 当a=b/2时候,所求式=(b^2-b^2/4)/(b^2/2)=3/2.