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解方程:(b-c)x^2+(c-a)x+(a-b)=0(b≠c)
人气:444 ℃ 时间:2020-05-13 03:30:18
解答
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
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