已知,a,b,c是△ABC的三边,求证:(a2+b2-c2)2-4a2b2<0.
已知,a,b,c是△ABC的三边,求证:(a2+b2-c2)2-4a2b2<0.
数学人气:693 ℃时间:2020-03-24 06:29:27
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证明:∵(a2+b2-c2)2-4a2b2=(a2+b2-c2)2-(2ab)2=(a2+b2-c2+2ab)(a2+b2-c2-2ab)=[(a2+2ab+b2)-c2][(a2-2ab+b2)-c2]=[(a+b)2-c2][(a-b)2-c2]=(a+b+c)(a+b-c)(a-b+c)(a-b-c),∵a,b,c是△A...
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