已知a,b,c满足a²+b²=c²,且a,b,c∈(0,+∞).(1)求证:log2[1+(b+c)/a]+log2[1+(a-c)/b]=1(2)设log4[1+(b+c)/a]=1,log8(a+b-c)=2/3,求a,b,c的值
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![已知a,b,c满足a²+b²=c²,且a,b,c∈(0,+∞).(1)求证:log2[1+(b+c)/a]+log2[1+(a-c)/b]=1(2)设log4[1+(b+c)/a]=1,log8(a+b-c)=2/3,求a,b,c的值](/uploads/image/z/5904512-8-2.jpg?t=%E5%B7%B2%E7%9F%A5a%2Cb%2Cc%E6%BB%A1%E8%B6%B3a%26%23178%3B%2Bb%26%23178%3B%3Dc%26%23178%3B%2C%E4%B8%94a%2Cb%2Cc%E2%88%88%EF%BC%880%2C%2B%E2%88%9E%EF%BC%89.%EF%BC%881%EF%BC%89%E6%B1%82%E8%AF%81%EF%BC%9Alog2%5B1%2B%28b%2Bc%29%2Fa%5D%2Blog2%5B1%2B%EF%BC%88a-c%EF%BC%89%2Fb%5D%3D1%282%29%E8%AE%BElog4%5B1%2B%28b%2Bc%29%2Fa%5D%3D1%2Clog8%28a%2Bb-c%29%3D2%2F3%2C%E6%B1%82a%2Cb%2Cc%E7%9A%84%E5%80%BC)
已知a,b,c满足a²+b²=c²,且a,b,c∈(0,+∞).(1)求证:log2[1+(b+c)/a]+log2[1+(a-c)/b]=1(2)设log4[1+(b+c)/a]=1,log8(a+b-c)=2/3,求a,b,c的值
已知a,b,c满足a²+b²=c²,且a,b,c∈(0,+∞).
(1)求证:log2[1+(b+c)/a]+log2[1+(a-c)/b]=1
(2)设log4[1+(b+c)/a]=1,log8(a+b-c)=2/3,求a,b,c的值
已知a,b,c满足a²+b²=c²,且a,b,c∈(0,+∞).(1)求证:log2[1+(b+c)/a]+log2[1+(a-c)/b]=1(2)设log4[1+(b+c)/a]=1,log8(a+b-c)=2/3,求a,b,c的值
答:
(1)等号左边
=log(2)(1+(b+c)/a+(a-c)/b+(ab+ac-bc-c^2)/ab)
=log(2)((ab+b^2+bc+a^2-ac+ab+ac-bc-c^2)/ab)
=log(2)((ab+bc-ac+ab+ac-bc)/ab)
=log(2)((2ab)/ab)
=log(2)2
=1=右边
所以原式成立
(2)
即(b+c)/a=3 ①,a+b-c=4 ②,
①得c=3a-b,②得c=a+b-4
所以3a-b=a+b-4,即a=b-2,代入①中得c=2b-6
由a^2+b^2=c^2得
(b-2)^2+b^2=(2b-6)^2
即b^2-10b+16=0,(b-8)(b-2)=0
所以b=8或b=2
当b=8时,a=6,c=10
当b=2时,a=0分母为0不符合.舍去.
所以a=6,b=8,c=10
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