Question

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

Answer 1

[(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