已知a分之1+b分之1=4,求2a+2b-7ab分之a+b-3ab的值.
问题描述：
已知a分之1+b分之1=4,求2a+2b-7ab分之a+b-3ab的值.
答
1
答
1/a +1/b =(a+b)/ab =4 => a+b=4ab
=> 2a+2b-7ab =2(a+b)-7ab=8ab-7ab=ab
=> a+b-3ab=4ab-3ab=ab => 所求= ab/ab =1.......ans
答
1/a+1/b=4
a+b=4ab
(a+b-3ab)/(2a+2b-7ab)
=(4ab-3ab)/(2*4ab-7ab)
=ab/(ab)
=1
答
因为1/a+1/b=4
通分得a+b=4ab
代入所求式，得
原式=4ab-3ab/2(4ab)-7ab
=1