已知1/a+1/b=2 则分式(3a-2ab+3b)/a+3ab+b)的值为?

问题描述:

已知1/a+1/b=2 则分式(3a-2ab+3b)/a+3ab+b)的值为?

∵1/a + 1/b = 2
∴通分得 (a + b)/ab = 2
∴a + b = 2ab
原式 = (3a - 2ab + 3b)/(a + 3ab + b)
={3(a + b) - 2ab}/(a + b + 3ab)
=(6ab - 2ab)/(2ab + 3ab)
=4/5

1/a +1/b=2两边同时乘以ab的a+b=2ab
方式(3a+3b-2ab)/(a+b+3ab)
=(3(a+b)-2ab)/((a+b)+3ab)
把a+b=2ab带入
=(3*(2ab)-2ab)/(2ab+3ab)
=4ab/5ab
=4/5

通分母:1/a+1/b=b/ab+a/ab=2
(b+a)/ab=2
2ab=a+b

(3a-2ab+3b)/(a+3ab+b)
=(3a-a-b+3b)/(a+3/2(a+b)+b)
=4/5

上下同时除以ab
(3a-2ab+3b)/a+3ab+b)
=(3/b-2+3/a)/(1/b+3+1/a)
=[3(1/a+1/b)-2]/[3+(1/a+1/b)]
=(6-2)/(3+2)
=4/5

分子分母同除以ab
(3a-2ab+3b)/a+3ab+b)
=(3/b+3/a-2)/(1/a+1/b+3)
=(6-2)/(2+3)
=4/5

因为1/a+1/b=2
所以a+b/ab=2
a+b=2ab
原式=[3(a+b)-2ab]/[(a+b)+3ab]
=(3*2ab-2ab)/(2ab+3ab)
=4ab/5ab
=4/5
=0.8