若log7 (2√2+1)+log2 (√2-1）=a,把log7 (2√2-1）+log2 (√2+1）表示为a的代数式

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若log7 (2√2+1)+log2 (√2-1）=a,把log7 (2√2-1）+log2 (√2+1）表示为a的代数式
若log7 (2√2+1)+log2 (√2-1）=a,把log7 (2√2-1）+log2 (√2+1）表示为a的代数式

若log7 (2√2+1)+log2 (√2-1）=a,把log7 (2√2-1）+log2 (√2+1）表示为a的代数式
两式相加
log7 (2√2+1)+log2 (√2-1）+log7 (2√2-1）+log2 (√2+1）
=log7 [(2√2+1)(2√2-1）] +log2 [(√2-1)(√2+1）]
=log7 7 + log2 1
=1+0=1
又log7 (2√2+1)+log2 (√2-1）=a
所以
log7 (2√2-1）+log2 (√2+1）=1-a

【解】：设b=log7 (2√2-1）+log2 (√2+1）;则有：
a+b=[log7 (2√2+1)+log2 (√2-1)]+[log7 (2√2-1）+log2 (√2+1）]
=[lg(2√2+1)/lg(7)+lg(2√2-1)/lg(7)]+[lg(√2-1)/lg(2)+lg(√2+1)/lg(2)]
=lg[(2√2+1))*(2√...

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