分数太少了,想想还是给你做
证明:
1.由 |lga| = |lgb|
得 lga = lgb 或 lga = -lgb
得 a = b 或 a = 1/b
因为 0 1 ,所以 b² > 1 ,所以 1/b² ∈ (0,1)
又 2|lg[(a+b)/2] = |lgb| ,b > 1 ,a = 1/b
则 2|lg[(1/b + b)/2]| = lgb
由基本不等式得 (1/b + b)/2 > 2/2 = 1
则 2lg[(1/b + b)/2] = lgb
即 [(1/b + b)/2]² = b
即 1/b² + 2 + b² = 4b
则 4b - b² = 2 + 1/b² ∈ (2,3)