求值:(lg^2)^3+(lg^5)^3+3lg^2*lg^5

发布网友发布时间：2024-10-01 17:45

共2个回答

热心网友时间：2024-10-09 04:40

(lg^2)^3+(lg^5)^3+3lg^2*lg^5
令a=lg2,b=lg5,则a+b=lg10=1
(lg^2)^3+(lg^5)^3+3lg^2*lg^5
=a^3+b^3+3ab
=(a+b)(a^2-ab+b^2)+3ab
=a^2-ab+b^2+3ab
=a^2+2ab+b^2
=(a+b)^2
=1

2^(1+1/2log2^5)
=2^(1+log2^5^(1/2))
=2^1*2^log2^5^(1/2))
=2*5^(1/2)
=2√5

7^(log7^6*log6^5*log5^4)
=[7^(log7^6)]^(log6^5*log5^4)
=6^(log6^5*log5^4)
=[6^(log6^5)]^(log5^4)
=5^(log5^4)
=4

1/3log1/6^8+2log1/6^√3
=log1/6^(8^1/3)+log1/6^(√3)^2
=log1/6^2+log1/6^3
=log1/6^6
=lg6/lg(1/6)
=lg6/lg(6^-1)
=lg6/(-lg6)
=-1

热心网友时间：2024-10-09 04:47

解:令lg2=x，lg5=y，x+y=lg10=1,x^3+y^3+3*xy=x^3+y^3+3xy(x+y)=(x+y)^3=1

热心网友时间：2024-10-09 04:43

热心网友时间：2024-10-09 04:41

解:令lg2=x，lg5=y，x+y=lg10=1,x^3+y^3+3*xy=x^3+y^3+3xy(x+y)=(x+y)^3=1