(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,
来源:学生作业帮助网 编辑:作业帮 时间:2024/10/03 22:18:22
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lg2(lg5-lg2)-lg5(lg5+3lg2)等于
(lg2)^3+(lg5)^3+3*lg2*lg5
(lg2)³+(lg5)³+3lg2×lg5
3+lg5+(lg2)^2
3lg2-lg5=
(lg2)^3+(lg5)^3+3lg2*lg5等于?
(lg2)^3+(lg5)^3+3lg2*lg5怎么算RT
(lg2)^3+3lg2·lg5+(lg5)^3 怎么算?
(lg2)^3+(lg5)^3+3lg2*lg5 这些怎么算.
(lg2)^3+3lg2·lg5+(lg5)^3 怎么算?
(lg2)^3+(lg5)^3+3lg2*lg5的值为?
(lg2)^3+3lg2乘lg5+(lg5)^3
求解3lg5+3lg2*lg5+3lg2^2~
(lg2)³+(lg5)³+3lg2×lg5(等于多少?)
(lg2)³+(lg5)³+3lg2×lg5(怎么算呀,
(lg2)³+(lg5)³+3lg2×lg5=?
(lg2)³+3lg2*lg5+(lg5)³等于多少
(lg²2-lg2*lg5+lg²)+3lg2*lg5