已知a/2=b/3≠0,求代数式:(5a-2b)/(a²-4b²)·(a-2b)的值

问题描述:

已知a/2=b/3≠0,求代数式:(5a-2b)/(a²-4b²)·(a-2b)的值

a/2=b/3
∴2b=3a
∴:(5a-2b)/(a²-4b²)·(a-2b)
=(5a-2b)/(a+2b)(a-2b)×(a-2b)
=(5a-2b)/(a+2b)
=(5a-3a)/(a+3a)
=2a/4a
=1/2从(5a-3a)/(a+3a)一步到2a/4a一步是怎么算的?上面的2b用3a代替 5a-3a=2aa+3a=4a