先化简,再求值:a^2+ab-ac/a^2-ab·(a-b)^2-c^2/a^2+2ab+b^2÷a^2-(b-c)^2/b^2-a^2.其中a=1,b=-2,c=-3
问题描述:
先化简,再求值:
a^2+ab-ac/a^2-ab·(a-b)^2-c^2/a^2+2ab+b^2÷a^2-(b-c)^2/b^2-a^2.
其中a=1,b=-2,c=-3
答
(a²+ab-ac)/(a²-ab)·[(a-b)²-c²]/(a²+2ab+b²)÷[a²-(b-c)²]/(b²-a²)
=a(a+b-c)/[a(a-b)]·(a-b-c)(a-b+c)/(a+b)²·(b-a)(b+a)/(a-b+c)(a+b-c)
=(b+c-a)/(a+b).
答
(a²+ab-ac)/(a²-ab)·[(a-b)²-c²]/(a²+2ab+b²)÷[a²-(b-c)²]/(b²-a²)=a(a+b-c)/[a(a-b)]·(a-b-c)(a-b+c)/(a+b)²...