natherain
花朵
共回答了17个问题采纳率:88.2% 举报
x-(a+b)x+ab=0 x-abx+(a+b)=0 ==> (x-a)(x-b) = 0 ==> (1-ab)x = -(a+b) ==> x =a,b ==> x = (a+b) / (ab-1) 假设 有 公共根,则有:a = (a+b) / (ab - 1) ==> a^2b - a = a+b ==> a^2b - 2a - b = 0 ==> a = [2 + √(4 +4b^2)] / 2b {a = [2 - √(4 +4b^2)] / 2b 舍去,因为 a>2} ==> a = [1 + √(1 + b^2)] / b a >2 ==> [1 + √(1 + b^2)] / b > 2 ==> 1 + √(1 + b^2) > 2b ==> √(1 + b^2) > 2b - 1 ==> 1 + b^2 > (2b-1)^2 ==> 1 +b^2 > 4b^2 - 4b + 1 ==> 3b^2 - 4b < 0 ==> b(3b - 4) < 0 ==> 4/3 < b < 0 与题干 a >2 相矛盾,所以 没有公共根.
1年前
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