Question

七年级数学，求解，必采纳，在线等，急急急急急急急急急急急急急急急

Answer 1

(a²+ab-ac)/(a²-ab)x[(a-b)²-c²]/(2ab+a²+b²)÷[a²-(b-c)²]/(a²-b²)
=[a(a+b-c)]/[a(a-b)]x[(a-b-c)(a-b+c)]/(a+b)²x[(a-b)(a+b)]/[(a-b+c)(a+b-c)]
=a-b-c
=

a=1 b=2 c=?????

追问

c=-3

追答

(a²+ab-ac)/(a²-ab)x[(a-b)²-c²]/(2ab+a²+b²)÷[a²-(b-c)²]/(a²-b²)
=[a(a+b-c)]/[a(a-b)]x[(a-b-c)(a-b+c)]/(a+b)²x[(a-b)(a+b)]/[(a-b+c)(a+b-c)]
=a-b-c
=1-2-（-3）
=-1+3
=2

Answer 2

Answer 3

追答