Question

初二方程计算题已知9a²-4b²=0,求代数式(b分之a)-(a分之b)-(ab分之a²+

Answer 1

9a²-4b²=0

(3a+2b)(3a-2b)=0

3a+2b=0,b/a=-3/2

或3a-2b=0,b/a=3/2

a/b-b/a-(a^2+b^2)/(ab)

=(a^2-b^2)/(ab)-(a^2+b^2)/(ab)

=(2b^2)/(ab)

=2b/a

=±3