一道高中数学题 高手进

已知log5(√6+1)+log2(√2-1)=a,则log5(√6-1)+log2(√2+1)=
2024-12-23 06:00:06
推荐回答(2个)
回答1:

设:log5(√6-1)+log2(√2+1)= b
a+b=log5(√6+1)+log2(√2-1)+log5(√6-1)+log2(√2+1)
=log5〔(√6+1)(√6-1)〕+log2〔(√2-1)(√2+1)〕
=log5(6-1)+log2(2-1)
=1+0=1
所以log5(√6-1)+log2(√2+1)= b=1-a

回答2:

注意到 (-1+√6)(√6+1)=6-1=5 => -1+√6=5/(1+√6)
(1+√2)(-1+√2)=2-1=1 => 1+√2 =1/(-1+√2)
log5(-1+√6)+log2(1+√2)
=log5[5/(1+√6)] + log2[1/(-1+√2)]
=log5(5)-log5(1+√6)-log2(-1+√2)
=1-a