设a、b、c为正数,且满足a2+b2=c2. (1)求证:log2(1+b+c/a)+log2(1+a−cb)=1 (2)若log4(1+b+c/a)=1,log8(a+b−c)=2/3,求a、b、c的值.

问题描述:

设a、b、c为正数,且满足a2+b2=c2
(1)求证:log2(1+

b+c
a
)+log2(1+
a−c
b
)=1
(2)若log4(1+
b+c
a
)=1
log8(a+b−c)=
2
3
,求a、b、c的值.

证明:(1)左边=log2

a+b+c
a
+log2
a+b−c
b
=log2(
a+b+c
a
a+b−c
b
)
=log2
(a+b)2c2
ab
=log2
a2+2ab+b2c2
ab
=log2
2ab+c2c2
ab
=log22=1

(2)由log4(1+
b+c
a
)=1
1+
b+c
a
=4
,∴-3a+b+c=0①
log8(a+b−c)=
2
3
a+b−c=8
2
3
=4

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.