高中题lg25+2/3lg8+lg20*lg5+(lg2)^2lg25+2/3lg8+lg20*lg5+(lg2)^2高中题．．．
问题描述：
高中题lg25+2/3lg8+lg20*lg5+(lg2)^2
lg25+2/3lg8+lg20*lg5+(lg2)^2
高中题．．．
答
3
设lg5=a，lg2=b。整理可得（a+b)(a+b+2), a+b=1
答
2lg5+2/33*lg2+(1+lg2)lg5+lg2*lg2=
2lg5+2lg2+lg5+lg2(lg5+lg2)=
2lg5+3lg2=
lg(5*5*2*2*2)=
lg200=
2+lg2
答
lg25+2/3lg8+lg20*lg5+(lg2)^2
=2lg5+2lg2+(lg4+lg5)lg5+(lg2)^2
=2lg5+2lg2+2lg2*lg5+(lg5)^2+(lg2)^2
=2(lg5+lg2)+(lg2+lg5)^2
=2lg10+(lg10)^2
=2+1
=3
答
=lg(25*（三次根号8）的平方)+lg2*lg10*lg5+lg2*lg2=lg100+lg2=10+lg2
答
lg25+2/3lg8+lg20*lg5+(lg2)^2
=2lg5+(2/3)*3lg2+lg(2*10)*lg5+(lg2)^2
=2lg5+2lg2+(lg2+lg10)*lg5+(lg2)^2
=2(lg2+lg5)+lg2*lg5+lg5+(lg2)^2
=2*1+lg2(lg5+lg2)+lg5
=2+lg2*1+lg5
=2+lg2+g5
=2+1
=3