www.5129.net > lg25+2/3lg8+lg5*lg20+(lg2)^2=?

lg25+2/3lg8+lg5*lg20+(lg2)^2=?

原式=2lg5+2lg2 +lg5(lg5+2lg2)+(lg2)^2 =2(lg5+lg2)+(lg2)^2+2lg2lg5+(lg5)^2 =2lg10+[lg2+lg5]^2 =2+1^2 =2+1 =3

lg5^2+2/3lg8+lg5*lg20+(lg2)^2 =2lg5+2/3lg2^3+lg5*lg5*2^2 +(lg2)^2 =2lg5+2/3 * 3 lg2+lg5(lg5+2lg2)+(lg2)^2 =2lg5+2lg2+(lg5)^2+2lg5lg2+(lg2)^2 =2(lg5+lg2)+(lg5)^2+2lg5lg2+(lg2)^2 =2lg10 + (lg5+lg2)^2 =2+(lg10)^2 =2+1=3

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lg25+2/3lg8+lg5lg20+lg2lg2 =lg5²+(2/3)lg2³+lg5(lg2*10)+lg2lg2 =2lg5+(2/3)*3lg2+lg5(lg2+lg10)+lg2lg2 =2lg5+2lg2+lg5(lg2+1)+lg2lg2 =2lg5+2lg2+lg5lg2+lg5+lg2lg2 =2(lg5+lg2)+lg5lg2+lg2lg2+lg5 =2(lg10)+lg2(lg5+lg2)+lg5 =2+...

(lg2)^2+lg25+2/3lg8+lg5*lg20 =(lg2)^2+2lg5+2lg2+lg5*(lg5+2lg2) =(lg2+lg5)^2+2(lg2+lg5) =(lg10)^2+2lg10 =1+2 =3

你有没有发现在第二步的后面三项(lg5)^2+2lg2*lg5+(lg2)^2 是完全平方,就是a^2+2ab+b^2=(a+b)^2类型的,具体点(lg5)^2+2lg2*lg5+(lg2)^2=(lg5+lg2)^2这是一个等式 至于第二步的第一项利用对数的运算法则就可以得出。

lg25+2/3lg8+lg5×lg20+(lg^2)2 =2lg5+2lg2+lg5*(lg5+2lg2)+(lg2)^2 =2(lg5+lg2)+(lg5)^2+2lg5*lg2+(lg2)^2 =2lg(2*5)+(lg5+lg2)^2 =2*1+1 =3

lg25+3分之2lg8+lg5*lg20+lg??2=lg5+lg5+(2/3)3lg2+lg5(2lg2+lg5)+lg^2 2=2lg5+2lg2+(2lg2lg5+lg^2 5+lg^2 2)=2(lg5+lg2)+(2lg2lg5+lg^2 5+lg^2 2)=2*1+(lg2+lg5)^2=2+1^2=2+1=3谢谢,这道题的关键在于对数的运算性质,结合完全平方公式。希望采...

=lg5²+2lg2+lg5lg20+lg²2 后面的一个应该是lg²2 =2lg5+2lg2+(1-lg2)(1+lg2)+lg²2 =2(lg5+lg2)+1-lg²2 +lg²2 =2+1 =3

(lg5)2+2/3lg8+lg5*lg20+(lg2)2 =(lg5)2+2lg2+lg5(lg5+2lg2)+(lg2)2 =(lg5)2+2lg5lg2+(lg2)2+2lg2+(lg5)2 =(lg5+lg2)2+2lg2+(lg5)2 =1+lg100lg2+(lg5)2 =1+(2lg5+lg2)lg2+(lg5)2 =1+(lg2)2+2lg5lg2+(lg5)2 =1+(lg2+lg5)2 =1+1 =2

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