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已知a/2=b/3=c/4,abc≠0,求2a^2-2bc+b^2/a^2-2bc-c^2
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已知a/2=b/3=c/4,abc≠0,求2a^2-2bc+b^2/a^2-2bc-c^2
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答案和解析
令a/2=b/3=c/4=t (t≠0),则a=2t,b=3t,c=4t
(2a²-2bc+b²)/(a²-2bc-c²)
=[2(2t)²-2×(2t)×(3t)+(3t)²]/[(2t)²-2×(3t)×(4t)-(4t)²]
=(5t²)/(-36t²)
=-5/36
(2a²-2bc+b²)/(a²-2bc-c²)
=[2(2t)²-2×(2t)×(3t)+(3t)²]/[(2t)²-2×(3t)×(4t)-(4t)²]
=(5t²)/(-36t²)
=-5/36
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