2lg5+(lg2)^3*(2lg5+lg2)计算~
来源：学生作业帮助网 编辑：六六作业网 时间：2024/05/09 09:29:41
2lg5+(lg2)^3*(2lg5+lg2)计算~2lg5+(lg2)^3*(2lg5+lg2)计算~2lg5+(lg2)^3*(2lg5+lg2)计算~设x=lg5,lg2=1-lg5=1-x2x
2lg5+(lg2)^3*(2lg5+lg2)计算~
2lg5+(lg2)^3*(2lg5+lg2)计算~
2lg5+(lg2)^3*(2lg5+lg2)计算~
设x=lg5,lg2=1-lg5=1-x
2x+(1-x)^3(1+x)
=2x+(1-x)^2(1-x^2)
=2x+(x^2-2x+1)(1-x^2)
=2x+x^2-x^4-2x+2x^3+1-x^2
=-x^4+2x^3+1
=x^3(2-x)+1
=(lg5)^3*lg20+1
好像只能化到这了,不知道了.
3+lg5+(lg2)^2
lg2(1+lg5)+(lg5)^2
计算：(lg5)^2 +lg2 *lg5
(lg5)^2+lg5.lg2=
求解3lg5+3lg2*lg5+3lg2^2~
2lg5+(lg2)^3*(2lg5+lg2)计算~
(lg²2-lg2*lg5+lg²)+3lg2*lg5
2lg2（lg2+lg5)+2lg5得数
(lg2)^2+lg2*lg5+(lg5)^2=
lg2*lg50+lg5*lg20-2lg5*lg2
(lg5)平方+2*lg5*lg2+(lg2)平方
lg5(1-lg5)+2lg5根号2+lg2(1-lg5)
(lg2)^3+(lg5)^2=
求值:(lg2)^3+3(lg2)^2*lg5+3lg2*(lg5)^2+(lg5)^3
(lg2)^3+(lg5)^3+lg5*lg8=(lg2+lg5)[(lg2)^2-lg2*lg5+(lg5)^2]+lg5*lg2^3=1[(lg2)^2-lg2*lg5+(lg5)^2]+3lg5*lg2=(lg2)^2+(lg5)^2+2lg5*lg2=(lg2+lg5)^2=1网上看了看答案,可是第一步没有看懂.求得出第一步的过程,
lg2+lg5=?1-lg2=?1-lg5=?2lg5+2/3lg8+lg5*lg20+(lg2)2=?
(lg2)^2+lg5*lg20
(lg2)^2+lg20*lg5