Question

已知∣3a-b+1∣+(3a-3/2b)²=0,求b²/a+b/(b/a-b*ab/a+b)的值

Answer 1

已知∣3a-b+1∣+(3a-3/2b)²=0,
则3a-b+1=0
3a-3/2b=0
a=-1
b=-2

b²/a+b/(b/a-b*ab/a+b)
=(a-b)/a
=1-b/a
=1-(-1)/(-2)
=1-1/2
=1/2

Answer 2

由∣3a-b+1∣+(3a-3/2b)²=0知
3a-b+1=0
3a-3/2b=0
从而解得b=-2, a=-1.
而b^2/a+b/(b/a-b*ab/a+b)=b^2/a+b/(b/a-b^2+b)
而b/a=2, 所以原式=-7/2
注：我觉得你后面的式好像缺少点东西，否则原式应该可以化得再简点，再代入值的。