一道计算和一道分式方程计算(c/a-b)^2/(-b/ac)^3/(-b/a)^4方程1/(x-2)+1/(x-6)=1/(x-7)+1/(x-1)半小时内回答加20

Question

一道计算和一道分式方程
计算 (c/a-b)^2/(-b/ac)^3/(-b/a)^4
方程 1/(x-2)+1/(x-6)=1/(x-7)+1/(x-1)
半小时内回答加20

Answer 1

(c/a-b)^2/(-b/ac)^3/(-b/a)^4
=-[(c-ab)/a]^2*(b/a)^4*c^3/(b/a)^3
=-[(c-ab)/a]^2bc^3/a
=-(ab-c)^2bc^3/a^3
1/(x-2)+1/(x-6)=1/(x-7)+1/(x-1)
(2x-8)/(x^2-8x+12)=(2x-8)/(x^2-8x+7)
分母恒不相等,分子相等,只有分子为0,方程才有意义.
2x-8=0
x=4