3倍lg2

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数学题lg5(lg8+lg1000)+(lg2倍根号3)^2+lg1/6+lg0.06

lg5(lg8+lg1000)+(lg2倍根号3)^2+lg1/6+lg0.06=lg5(lg2^3+lg10^3)+((lg2)*(根号3))^2+lg(1/6)*0.06=lg5(3lg2+3)+

-2倍的lg5*-2倍的lg3*-3倍的lg2等于多少?对数乘法公式是什么,

=-2(lg5*2)-lg2-2lg3=-2*1-lg(2/9)=-2-.弱弱的说一句----这题你是不是记错了啊,算不出来啊.再问:谢了,这题是最简的,不能算了。我也是刚知道,真不好意思!!谢了

(lg2)^3+(lg5)^3+3lg2*lg5等于?

1(lg2+lg5)(lg2^2-lg2*lg5+lg5^2)+3lg2*lg5=(lg2^2-lg2*lg5+lg5^2)+3lg2*lg5=(lg2+lg5)^2=1努力吧,把对数函数的规律记记清

3lg2*lg5+3(lg2)的平方-3,计算,

3lg2*lg5+3(lg2)^2(提取公因式)=3lg2(lg5+lg2)=3lg2

(lg²2-lg2*lg5+lg²)+3lg2*lg5

括号中最后lg写掉了数字,当他是5(lg²2-lg2*lg5+lg²5)+3lg2*lg5=lg²2+lg²5-lg2*lg5+3lg2*lg5=lg²

(lg2)^3+(lg5)^3+3lg2*lg5 这些怎么算.

令a=lg2,b=lg5则a+b=lg2+lg5=1原式=a³+b³+3ab=(a+b)(a²-ab+b²)+3ab=a²-ab+b²+3a

(lg3/2lg2+lg3/3lg2)*(lg2/lg3+lg2/2lg3)

(lg3/2lg2+lg3/3lg2)*(lg2/lg3+lg2/2lg3)=【log(4)3+log(8)3】*【log(3)2+log(9)2】=【1/2log(2)3+1/3log(2)3】*【

3+lg5+(lg2)^2

原式=2lg5+(2/3)lg8+lg5*lg20+(lg2)^2=2lg5+2lg2+lg5*(lg5+2lg2)+(lg2)^2=2lg5+2lg2+(lg5)^2+2lg2lg5+(lg2)^2

(lg2)^3+(lg5)^3+3*lg2*lg5

(lg2)^3+(lg5)^3+3*lg2*lg5前面用立方和=(lg2+lg5)[(lg2)^2-lg2*lg5+(lg5)^2]+3*lg2*lg5=lg10[(lg2)^2-lg2*lg5+(l

(lg2)³+(lg5)³+3lg2×lg5

(lg2)³+(lg5)³+3lg2×lg5=(lg2)³+(lg5)³+3lg2×lg5×lg10=(lg2)³+(lg5)³+3lg2×

(lg2)^3+3lg2乘lg5+(lg5)^3

(lg2)^3+3lg2*lg5+(lg5)^3=(lg^32+lg^35)+3lg2*lg5=(lg2+lg5)(lg^22-lg2*lg5+lg^25)+3lg2*lg5=(lg^22-lg2*l

(lg3分之2lg2+lg3分之3lg2)(lg2分之lg3+lg2分之2lg3)

再问:第一步不懂为什么lg3+lg3=5lg3lg2+lg2=3lg2再答:分母通分就可以了。再问:soga谢谢

(lg2)^3+3lg2·lg5+(lg5)^3 怎么算?

(lg2)^3+3lg2·lg5+(lg5)^3=(lg2)^3+(lg5)^3+3lg2·lg5=(lg2+lg5)[(lg2)^2+(lg5)^2-lg2lg5]+3lg2·lg5=[(lg2+l

lg2(lg5-lg2)-lg5(lg5+3lg2)等于

lg2(lg5-lg2)-lg5(lg5+3lg2)=lg2lg5-(lg2)^2-(lg5)^2-3lg5lg2=-[(lg2)^2+2lg2lg5+(lg5)^2]=-(lg2+lg5)^2=-[

(lg2)^3+(lg5)^3+3lg2*lg5怎么算

=(lg2+lg5)((lg2)^2-lg2*lg5+(lg5)^2)+3lg2*lg5=lg(2*5)((lg2)^2-lg2*lg5+(lg5)^2)+3lg2*lg5=(lg2)^2-lg2*l

lg5(lg8+lg1000)+(lg2倍根3)^2+lg1/6+lg0.06要详细步骤

lg5(lg8+lg1000)+(lg2倍根号3)^2+lg1/6+lg0.06=lg5(lg2^3+lg10^3)+((lg2)*(根号3))^2+lg(1/6)*0.06=lg5(3lg2+3)+

(lg2)^3+(lg5)^3+3lg2*lg5的值为?

设x=lg2,y=lg5,则x+y=lg2+lg5=lg10=1原式=x^3+y^3+3xy=(x+y)(x^2-xy+y^2)+3xy=x^2-xy+y^2+3xy=x^2+2xy+y^2=(x+y

3*(lg2)^2+3(1-lg2)*lg2+2(1-lg2)-lg2+9/2

3*(lg2)^2+3(1-lg2)*lg2+2(1-lg2)-lg2+9/2=3*(lg2)^2+3lg2-3(lg2)^2+2-2lg2-lg2+9/2=2+9/213/2

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

1.设x=lg2,y=lg5,则x+y=lg2+lg5=lg10=1原式=x^3+y^3+3xy=(x+y)(x^2-xy+y^2)+3xy=x^2-xy+y^2+3xy=x^2+2xy+y^2=(x

lg2

解题思路:对a分类讨论解题过程:varSWOC={};SWOC.tip=false;try{SWOCX2.OpenFile("http://dayi.prcedu.com/include/readq.