问题描述: 化简后求值[(a+3)/(a^2-4)+a/(6-a-a^2)]除以(4a+9)/(5a-10),其中a=-8 1个回答 分类:数学 2014-12-15 问题解答: 我来补答 [(a+3)/(a^2-4)+a/(6-a-a^2)]除以(4a+9)/(5a-10)先化简= [(a+3)/(a+2)(a-2)]-a/(a+3)(a-2)除以(4a+9)/(5a-10)通分 (a+2)(a-2)(a+3)=[(a+3)^2 - a(a+2)]/(a+2)(a-2)(a+3)除以(4a+9)/(5a-10)先化简,再约分 =( a^2+6a+9-a^2-2a)/(a+2)(a-2)(a+3)除以(4a+9)/(5a-10)=[4a+9/(a+2)(a-2)(a+3)]/(4a+9)/(5a-10)约掉4a+9 ,a-2 =[1/(a+2)(a+3)]x5把a = -8 所以等式= 1/6 展开全文阅读