www.5129.net > lg5+lg2*lg5+(lg2)^2=

lg5+lg2*lg5+(lg2)^2=

(lg2)^2=lg2*lg2 (lg2)^2+lg2*lg5+lg5( =lg2*lg2+lg2*lg5+lg5(前两项提取公因式lg2) =lg2(lg2+lg5)+lg5 =lg2*1+lg5 =1

lg5^2+2lg2-lg2^2 =2lg5+2lg2-2lg2 =2lg5; (lg5)^2+2lg2-(lg2)^2? =(lg5)^2-(lg2)^2+2lg2 =(lg5+lg2)(lg5-lg2)+lg2² =lg10*lg5/2+lg4 =1*lg5/2+lg4 =lg[(5/2)*4] =lg10 =1

lg5^2+2/3lg8+lg5*lg20+(lg2)^2 =2lg5+2/3lg2^3+lg5*lg5*2^2 +(lg2)^2 =2lg5+2/3 * 3 lg2+lg5(lg5+2lg2)+(lg2)^2 =2lg5+2lg2+(lg5)^2+2lg5lg2+(lg2)^2 =2(lg5+lg2)+(lg5)^2+2lg5lg2+(lg2)^2 =2lg10 + (lg5+lg2)^2 =2+(lg10)^2 =2+1=3

lg2*lg5 =lg(2*5) =lg10 =1 lg^25=lg5^2

完成这个问题之前,首先要对对数的计算规则有所了解。 lgm+lgn=lg(mn lgm-lgn=lg(m/n mlgn=lg(n^m) 因此lg2*+lg5=lg10=1, 但是lg2*lg5就无法在进行合并化简了。

由公式:lga+lgb=lgab; 则,lg5+lg2=lg(5×2)=lg10=1 (由对数函数定义及其与指数函数互为反函数的关系:10^1=10→lg10=1)

不是,指数运算才是底数不变指数相加,对数应该是同底数相加底数相乘。

(lg2)² +lg5 +lg5lg2 =(lg2)² +lg5(1+lg2) = (lg2)² +lg5(lg10+lg2) =(lg2)² +lg5(lg20) =(lg2)² +lg5(2lg2+lg5) =(lg2)^2 + 2lg2lg5 + (lg5)^2 =(lg2+lg5)^2 =1

(lg5)^2+lg2*lg50 =(lg5)^2+lg2*(1+lg5) = (lg5)^2+lg2*lg5+lg2 =lg5(lg5+lg2)+lg2 =lg5+lg2 =1

=lg5²+2lg2+lg5lg20+lg²2 后面的一个应该是lg²2 =2lg5+2lg2+(1-lg2)(1+lg2)+lg²2 =2(lg5+lg2)+1-lg²2 +lg²2 =2+1 =3

网站地图

All rights reserved Powered by www.5129.net

copyright ©right 2010-2021。
www.5129.net内容来自网络,如有侵犯请联系客服。zhit325@qq.com