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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【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√2-1)]/lg(7)+lg[(√2-1)*(√2+1)]/lg(2)
=lg(7)/lg(7)+lg(1)=1;
所以:b=1-a
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