问题
解答题
设a、b、c为正数,且满足a2+b2=c2. (1)求证:log2(1+
(2)若log4(1+
|
答案
证明:(1)左边=log2
+log2a+b+c a
=log2(a+b-c b
•a+b+c a
)a+b-c b
=log2
=log2(a+b)2-c2 ab
=log2a2+2ab+b2-c2 ab
=log22=1;2ab+c2-c2 ab
(2)由log4(1+
)=1得1+b+c a
=4,∴-3a+b+c=0①b+c a
由log8(a+b-c)=
得a+b-c=82 3
=4②2 3
由①+②得b-a=2③
由①得c=3a-b,代入a2+b2=c2得2a(4a-3b)=0,∵a>0,
∴4a-3b=0④
由③、④解得a=6,b=8,从而c=10.