来自秦政的问题
(log2)^2-log5*log20课本上等于(log2)^2+(1+log2)(1-log2)=1;(log2)^2+(1+log2)(1-log2)是怎么来的?
(log2)^2-log5*log20
课本上等于(log2)^2+(1+log2)(1-log2)=1;
(log2)^2+(1+log2)(1-log2)是怎么来的?


(log2)^2-log5*log20课本上等于(log2)^2+(1+log2)(1-log2)=1;(log2)^2+(1+log2)(1-log2)是怎么来的?
(log2)^2-log5*log20
课本上等于(log2)^2+(1+log2)(1-log2)=1;
(log2)^2+(1+log2)(1-log2)是怎么来的?
【【注:好像你的题目中的对数是,以10为底的常用对数.】】lg5=lg(10/2)=(lg10)-(lg2)=1-(lg2)lg20=lg(2×10)=(lg2)+(lg10)=1+(lg2)∴(lg5)×(lg20)=(1-lg2)(1+lg2)=1-(lg2)²∴(lg2)²+(lg5)(lg20)=1...