高一对数运算，谢！
1、（lg根27+lg8-lg根1000）/(lg1.2)2、（1/2）lg（32/49）-（4/3）lg根8+lg根245...
1、（lg根27+lg8-lg根1000）/(lg1.2)
2、（1/2）lg（32/49）-（4/3）lg根8+lg根245 展开
2、（1/2）lg（32/49）-（4/3）lg根8+lg根245 展开
展开全部
1、(lg根27+lg8-lg根1000）/(lg1.2)
=[lg3^(3/2)+lg2^3-lg10^(3/2)]/(lg1.2)
=[lg3^(3/2)+lg4^(3/2)-lg10^(3/2)]/(lg1.2)
=(3/2)(lg3+lg4-lg10)/(lg1.2)
=(3/2)lg(3*4/10)/lg1.2
=(3/2)lg1.2/lg1.2
=3/2
2、(1/2)lg（32/49）-（4/3）lg根8+lg根245
=(1/2)lg(2^5/7^2)-(4/3)lg2^(3/2)+lg(5*7^2)^(1/2)
=(1/2)lg(2^5/7^2)-(4/3)*(3/2)lg2+(1/2)*lg(5*7^2)
=(1/2)[lg(2^5/7^2)+lg(5*7^2)]-2lg2
=(1/2)lg(2^5*5)-2lg2
=(1/2)lg(2^4*10)-2lg2
=(1/2)lg2^4+(1/2)*lg10-2lg2
=2lg2+1/2-2lg2
=1/2
=[lg3^(3/2)+lg2^3-lg10^(3/2)]/(lg1.2)
=[lg3^(3/2)+lg4^(3/2)-lg10^(3/2)]/(lg1.2)
=(3/2)(lg3+lg4-lg10)/(lg1.2)
=(3/2)lg(3*4/10)/lg1.2
=(3/2)lg1.2/lg1.2
=3/2
2、(1/2)lg（32/49）-（4/3）lg根8+lg根245
=(1/2)lg(2^5/7^2)-(4/3)lg2^(3/2)+lg(5*7^2)^(1/2)
=(1/2)lg(2^5/7^2)-(4/3)*(3/2)lg2+(1/2)*lg(5*7^2)
=(1/2)[lg(2^5/7^2)+lg(5*7^2)]-2lg2
=(1/2)lg(2^5*5)-2lg2
=(1/2)lg(2^4*10)-2lg2
=(1/2)lg2^4+(1/2)*lg10-2lg2
=2lg2+1/2-2lg2
=1/2
本回答由提问者推荐
已赞过
已踩过<
评论
收起
你对这个回答的评价是？
推荐律师服务：
若未解决您的问题，请您详细描述您的问题，通过百度律临进行免费专业咨询
