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=lg1
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
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已知a+b=(1g2)^3+(1g2)^5+31g2*1g5,求3ab+a^3+b^3的值