已知log7(2√2+1)+log2(√2-1)=a,则log7(2√2-1)+log2(√2+1)=_______.
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已知log7(2√2+1)+log2(√2-1)=a,则log7(2√2-1)+log2(√2+1)=_______.已知log7(2√2+1)+log2(√2-1)=a,则log7(2√2-1)+l
已知log7(2√2+1)+log2(√2-1)=a,则log7(2√2-1)+log2(√2+1)=_______.
已知log7(2√2+1)+log2(√2-1)=a,则log7(2√2-1)+log2(√2+1)=_______.
已知log7(2√2+1)+log2(√2-1)=a,则log7(2√2-1)+log2(√2+1)=_______.
利用分子有理化,分子分母同乘以分子的有理化因子,向已知条件转化.
log7(2√2-1)+log2(√2+1)
=log7 [(8-1)/(2√2+1)] + log2 [(2-1)/(√2-1)]
=1- log7 (2√2+1) - log2 (√2-1)
=1- [log7(2√2+1)+log2(√2-1)]
=1-a.
设log7(2√2-1)+log2(√2+1)=X 则a+X=log7(2√2+1)(2√2-1)+log2(√2-1)(√2+1)=1+0=1 则X=1-a log7(2√2-1)+log2(√2+1)=1-a
1-a
log7(2√2-1)+log2(√2+1)
=log7[7/(2√2+1))+log2[1/(√2-1)]
=1-log7(2√2+1)-log2(√2-1)
=1-a.
已知log7(2√2+1)+log2(√2-1)=a,则log7(2√2-1)+log2(√2+1)=_______.
(log5 √2*log7 9)/(log5 1/3 *log7 3√4)+log2( √(3+√5)- log√(3-√5))=?
数学(log5 √2*log7 9)/(log5 1/3 *log7 3√4)+log2( √(3+√5)- log√(3-√5))
数学(log5 √2*log7 9)/(log5 1/3 *log7 3√4)+log2( √(3+√5)- log√(3-√5))
已知log7[log3(log2 x)]=0,求x^(-1/2)
已知log7【log3(log2 X)】=0,那么x^(-1/2)等于?
已知log7[log3(log2∧x)]=0,那么x∧-1/2等于
已知log7[log3(log2 x)]=0,求x^(-1/2)
已知log7[log3(log2 x)]=0,求x^(-1/2)
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
求下列函数的定义域(1)y=1/log2 X(2) y=log7(1/1-3x)(3) y=√log3 X
3^log9(25)+16^log4(2)-(根号7)^log7(9)+(-1)^log2(1)
如果 log7[log3(log2^x)]=0 那么x^1/2等于没人答=.
2log5(底数) 25+3log2(底数) 64-8log7 (底数)1的值为是多少,
求下列函数的定义域 (1)y=1/log2 X (2) y=log7(1/1-3x) (3) y=求下列函数的定义域(1)y=1/log2 X(2) y=log7(1/1-3x)(3) y=√log3 X:
√(㏒7 2的平方-log7 4+1)化简
(log5根号2乘以log7 9)除(log5 3分之1乘log7 3倍的根号4)+log2(根号(3+根号5)—根(3-根号5))
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