解方程：(b-c)x^2+(c-a)x+(a-b)=0(b≠c)
问题描述：
解方程：(b-c)x^2+(c-a)x+(a-b)=0(b≠c)
答
Δ = (c-a)^2 - 4(b-c)(a-b)= c^2 - 2ac + a^2 - 4ab + 4b^2 + 4ac - 4bc= (a+c)^2 - 4(a+c)b + 4b^2= (a+c-2b)^2x = [ -(c-a) ± √Δ ] / 2(b-c)= [ a - c ± (a+c-2b) ] / 2(b-c)= (2a - 2b) / 2(b-c) = (a-b)/(b...最后一个答案错了吧 a-c-(a+c-2b)去括号后应该是+2b吧嗯，是 +2b，最终答案是：x1 = (a-b)/(b-c)x2 = 1