已知1/a+1/b=4则a-3ab+b/2a+2b-7ab=?
问题描述：
已知1/a+1/b=4则a-3ab+b/2a+2b-7ab=?
一道2007年的赤峰题
答
a-3ab+b/2a+2b-7ab
分子分母同除以ab
=(a/ab-3ab/ab+b/ab)/(2a/ab+2b/ab-7ab/ab)
=(1/b-3+1/a)+(2/b+2/a-7)
=[(1/a+1/b)-3]/[2(1/a+1/b)-7]
=(4-3)/(8-7)
=1