(lg5)^2+(lg2)(lg50)
= (lg5)^2+(lg2)(lg25×2)
= (lg5)^2+(lg2)[lg(5^2)+lg2]
= (lg5)^2+(lg2)(2lg5+lg2)
= (lg5)^2+2(lg5)(lg2)+(lg2)^2
= (lg5+lg2)^2
= [lg(5×2)]^2
= (lg10)^2
= 1^2
= 1
log3*(√3/3)+log8*(4)
= [lg(√3/3)]/lg3+lg4/lg8
= (lg√3-lg3)/lg3+(lg2^2)/(lg2^3)
= [lg3^(1/2)-lg3]/lg3+(2lg2)/(3lg2)
= [(1/2)lg3-lg3]/lg3+2/3
= (1/2)-1+2/3
= 2/3-1/2
= 1/6
4^[log2*(5)]+2×5^[log25*(4)]
= (2^2)^[log2*(5)]+2×[25^(1/2)]^[log25*(4)]
= 2^2[log2*(5)]+2×25^(1/2)[log25*(4)]
= {2^[log2*(5)]}^2+2×{25^[log25*(4)]}^(1/2)
= 5^2+2×4^(1/2)
= 25+2×2
= 29
寫得夠清楚了吧同學,因為我實在是無聊,已經(jīng)很晚了……