(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?

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(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?
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(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

没能够理解你要问的是什么

log5=log(10/2)=log10-log2=1-log2
log20=log2*10=log2+log10=1+log2
∴代入就行了。

(1+log2)(1-log2) =1-log2*1+1*log2-(log2)^2=1-(log2)^2
（log2）^2 +(1+log2)(1-log2) =(log2)^2+1-(log2)^2=1

将(1+log2)(1-log2) 化简可得
=1×1-1×log2+1×log2-log2×log2
=1-(log2)^2
（log2）^2 +(1+log2)(1-log2)
=（log2）^2+1-(log2)^2
=1

怎样计算（log2)^2+(log5)^2+(2*log2)*log5 (log5)^2+(log2)(log50)=? (log5)^2+log2·log50 (log2)^2+(log5)^2+log2*log25 log2^5乘以log5^2等于多少? (log3^4+log5^2)(log2^9+log2^根号3) (log5(10))^3-(log5(2))^3-3log5(10)*log2(5) lg(1/4)-lg2(5) 2log5(10)+log5(0.25) 2log2(25)-3log2(64) log2(log2(16) ) log2/log5等于多少 （log5）^2+2lg2-（log2）^2 计算 log2^25*log3^2倍根号2*log5^9= log2^5*log3^2倍根号2*log5^9= 2^(1/2log2 3)+(1/25)^log5 3= log3 25×log2 27×log5 2^-3=_____ 求(log2 5+log4 0.2)(log5 2+log25 0.5) 求值:log2∧5×log5∧1/2 (log2^5+log4^0.2)(log5^2+log25^0.25)化简 2log5(25)-3log2(64) 如题`怎么做?