Question

初二数学题分式化简

化简:[a²(1/b-1/c)+b²(1/c-1/a)+c²(1/a-1/b)]/[a(1/b-1/c)+b(1/c-1/a)+c(1/a-1/b)]，要详细过程，麻烦了。

Answer 1

原式=[a^3(c-b)+b^3(a-c)+c^3(b-a)]/[a^2(c-b)+b^2(a-c)+c^2(b-a)] （上下同乘abc）
=[a^3(c-b)+a(b^3-c^3)+bc(c^2-b^2)]/a^2(c-b)+a(b^2-c^2)+bc(c-b)]
={(c-b)[a^3-a(b^2+bc+c^2)+bc(b+c)]}/{(c-b)[a^2-a(b+c)+bc}
=[a(a^2-b^2)-av(b+c)+bc(b+c)]/(a^2-ab-ac+bc)
=[a(a+b)(a-b)-c(b+c)(a-b)]/[(a-b)(a-c)]
=[a(a+b)-c(b+c)]/(a-c)
=[a(a+b)+ac-ac-c(b+c)]/(a-c)
=[a(a+b+c)-c(a+b+c)]/(a-c)
=[(a-c)(a+b+c)]/(a-c)
=a+b+c

Answer 2

=====a + b + c (直接称乘出来)

Answer 3

=[(c-b)a3+(a-c)b3+(b-a)c3]/[(c-b)a2+(a-c)b2+(b-a)c2]