解(1g2)^3+(1g5)^3+1g5*1g8

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解(1g2)^3+(1g5)^3+1g5*1g8
解(1g2)^3+(1g5)^3+1g5*1g8

解(1g2)^3+(1g5)^3+1g5*1g8
令x=lg2,y=lg5,x+y=lg2+lg5=lg10=1
x³+y³+y*(3x)
=(x+y)³-3xy(x+y)+3xy
=1³-3xy+3xy
=1

=(lg2)^3+(lg5)^3+3lg5*lg2=(lg2)^3+(lg5)^3+3lg5*lg2*1=(lg2)^3+(lg5)^3+3lg5*lg2*(lg2+lg5)=(lg2)^3+(lg5)^3+3(lg2)^2*(lg5)+3(lg5)^2=(lg2+lg5)^3=1

解(1g2)^3+(1g5)^3+1g5*1g8 化简求值 (1g2)^3+(1g5)^3+3*1g2*1g5 (1g2)^3+(1g5)^3+31g2*1g5=? 1g2+1g5= 1g2+1g5等于 1g5^2+1g8^(2/3)+1g5·1g20+(1g2)^2= (1g5)+21g2-(1g2) 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 (1g2)的二次方+1g20乘以1g5 计算(1g5)2+1g2*1g50=多少? 计算2sin180°-1g5-1g2?有急用对数计算（1g2）²+1g5·lg20 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