(1g2)^3+(1g5)^3+31g2*1g5=?
来源：学生作业帮助网 编辑：六六作业网 时间：2024/11/08 00:06:35
(1g2)^3+(1g5)^3+31g2*1g5=?(1g2)^3+(1g5)^3+31g2*1g5=?(1g2)^3+(1g5)^3+31g2*1g5=?(1g2)^3+(1g5)^3+31g2*1
(1g2)^3+(1g5)^3+31g2*1g5=?
(1g2)^3+(1g5)^3+31g2*1g5=?
(1g2)^3+(1g5)^3+31g2*1g5=?
(1g2)^3+(1g5)^3+31g2*1g5
＝(lg2+lg5)[(lg2)^2+(lg5)^2-lg2lg5]+31g2*1g5
=[(lg2)^2+(lg5)^2-lg2lg5]+31g2*1g5
=[(lg2)^2+(lg5)^2+2lg2lg5]
=(lg2+lg5)^2
=1
(1g2)^3+(1g5)^3+31g2*1g5
=(lg2+lg5)[(1g2)^2+(1g5)^2-1g2*1g5]+31g2*1g5
=[(1g2)^2+(1g5)^2-1g2*1g5]+31g2*1g5
=(1g2)^2+(1g5)^2+21g2*1g5
=(lg2+lg5)^2
=1^2
=1
(1g2)^3+(1g5)^3+31g2*1g5=?
化简求值 (1g2)^3+(1g5)^3+3*1g2*1g5
(1g5)+21g2-(1g2)
1g2+1g5=
1g2+1g5等于
已知a+b=(1g2)^3+(1g2)^5+31g2*1g5,求3ab+a^3+b^3的值
已知a+b=(1g2)^3+(1g2)^5+31g2*1g5,求3ab+a^3+b^3的值
解(1g2)^3+(1g5)^3+1g5*1g8
1g5^2+1g8^(2/3)+1g5·1g20+(1g2)^2=
lg5+lg2*1g5+(1g2)^2
1g5-1g2份之一 计算
1g5+1g2等于多少.
1g2+1g5+(-8)的-1/3次方 过程 在此可以跟我解释一下,1g2+1g5怎么算
1g25+2/3g8+1g5乘1g20+(1g2)的平方
(1g5)的2平方+21g2-(1g2)的2平方
( 1g5 )^2+21g2-( 1g2)^2计算此题.
(1g2)的二次方+1g20乘以1g5
计算(1g5)2+1g2*1g50=多少?