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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

(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

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...

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

lg25+2/3lg8+lg20*lg5+(lg2)^2 =2lg5+(2/3)*3lg2+lg(2*10)*lg5+(lg2)^2 =2lg5+2lg2+(lg2+lg10)*lg5+(lg2)^2 =2(lg2+lg5)+lg2*lg5+lg5+(lg2)^2 =2*1+lg2(lg5+lg2)+lg5 =2+lg2*1+lg5 =2+lg2+g5 =2+1 =3

(1)原式=sin60°?cos30°+sin30°?cos60°=sin(30 0 +60 0 )=sin90°=1.(2)原式=2lg5+2lg2+lg5?(2lg2+lg5)+lg2?lg2=2+(lg5) 2 +2lg2?lg5+(lg2) 2 =2+(lg5+lg2) 2 =3. (3)原式= 2? 3 1 2 ? 3 1 3 ? 2 - 1 3 ? 3 1 6 ? 2 1 3 = 2 1-...

=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

lg25+2÷3×lg8=2lg5+2lg2=2lg10=2 lg5×lg20+(lg2)²=lg5×(2lg2+lg5)+(lg2)²=(lg5)²+2lg5×lg2+(lg2)2=(lg2+lg5)²=1 所以得3

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