1.计算:(1+(2^(-1/8))(二的负八分之一次幂))(1+(2^(-1/4)))(1+(2^(-1/2)))2.已知f(x)=x^a,且f(x)的图像过点(8,1/4),求f(27)的值
来源:学生作业帮助网 编辑:六六作业网 时间:2025/02/01 22:58:49
1.计算:(1+(2^(-1/8))(二的负八分之一次幂))(1+(2^(-1/4)))(1+(2^(-1/2)))2.已知f(x)=x^a,且f(x)的图像过点(8,1/4),求f(27)的值
1.计算:(1+(2^(-1/8))(二的负八分之一次幂))(1+(2^(-1/4)))(1+(2^(-1/2)))
2.已知f(x)=x^a,且f(x)的图像过点(8,1/4),求f(27)的值
1.计算:(1+(2^(-1/8))(二的负八分之一次幂))(1+(2^(-1/4)))(1+(2^(-1/2)))2.已知f(x)=x^a,且f(x)的图像过点(8,1/4),求f(27)的值
1
(1+(2^(-1/8))(二的负八分之一次幂))(1+(2^(-1/4)))(1+(2^(-1/2)))
=(1+(2^(-1/8))(二的负八分之一次幂))(1+(2^(-1/4)))(1+(2^(-1/2))) *(1-(2^(-1/8))/(1-(2^(-1/8))
=[1-2^(-1)]/[1-2^(-1/8)]
=1/[2-2^(7/8)]
2
f(x)=x^a
1/4=8^a
2^(-2)=2^(3a)
3a=-2
a=-2/3
f(x)=x^(-2/3)
f(27)=27^(-2/3)=3^(-3*2/3)=3^(-2)=1/9
(1+(2^(-1/8))(1+(2^(-1/4)))(1+(2^(-1/2)))
= (1-(2^(-1/8))(1+(2^(-1/8))(1+(2^(-1/4)))(1+(2^(-1/2)))/(1-(2^(-1/8)))
=(1-(2^(-1/4))(1+(2^(-1/4)))(1-(2^(-1/2))^2)/(1-(2^(-1/8)))
=(1-(2^(-1/...
全部展开
(1+(2^(-1/8))(1+(2^(-1/4)))(1+(2^(-1/2)))
= (1-(2^(-1/8))(1+(2^(-1/8))(1+(2^(-1/4)))(1+(2^(-1/2)))/(1-(2^(-1/8)))
=(1-(2^(-1/4))(1+(2^(-1/4)))(1-(2^(-1/2))^2)/(1-(2^(-1/8)))
=(1-(2^(-1/2))(1-(2^(-1/2))/(1-(2^(-1/8)))
=(1-2^(-1))/(1-2^(-1/8))
=(1/2)/(1-2^(-1/8))
=1/(2-2^(1-1/8))
=1/(2-2^(7/8))
f(x)的图像过点(8,1/4),
所以1/4=8^a
2^(-2)=2^(3a)
3a=-2
a=-2/3
f(x)=x^(-2/3)
f(27)=27^(-2/3)=(3^3)^(-2/3)=3^(-2)=1/9
收起
1.设a=2^(-1/8),则a^8=1/2
原式=(1+a)a(1+a^2)(1+a^4)=a(1-a^8)/(1-a)=a/(2(1-a))=1/(2(1/a-1))=1/(2(2^(1/8)-1))
2.即1/4=8^a,2^(-2)=2^(3a)。所以-2=3a,a=-2/3
f(27)=27^(-2/3)=1/9
1.
(1-(2^(-1/8))*(1+(2^(-1/8))(1+(2^(-1/4)))(1+(2^(-1/2))) /(1-(2^(-1/8))
=(1-(2^(-1/4)))*(1+(2^(-1/4)))(1+(2^(-1/2))) /(1-(2^(-1/8))
=……
=(1-2^(-1))/(1-(2^(-1/8))
=1/(2-2^(7/8))
全部展开
1.
(1-(2^(-1/8))*(1+(2^(-1/8))(1+(2^(-1/4)))(1+(2^(-1/2))) /(1-(2^(-1/8))
=(1-(2^(-1/4)))*(1+(2^(-1/4)))(1+(2^(-1/2))) /(1-(2^(-1/8))
=……
=(1-2^(-1))/(1-(2^(-1/8))
=1/(2-2^(7/8))
2.
f(8)=8^a=1/4
(2^3)^a=2^(-2)
=>
3a=-2
a=-2/3
f(27)
=27^(-2/3)
=3^(3*(-2/3))
=1/9
收起