简单一道对数计算题

看到5，2，lg想到10和完全平方
=(lg5)^2 +lg2*(lg25+lg2)其中lg25=2lg5

=(lg5)^2+2lg2×lg5+(lg2)^2
=(lg5+lg2)^2
=1
lgx+lgy=lg(xy)

(lg5)^2 + lg2*lg50 = (lg5)^2 + lg2*lg（25*2）
= (lg5)^2 + lg2*(lg25+lg2)
= (lg5)^2 + lg2*lg5^2+(lg2)^2
= (lg5)^2 + 2lg2*lg5+(lg2)^2
=(lg5+lg2)^2
=[lg(5*2)]^2
=(lg10)^2
=1

(lg5)^2 + lg2*lg50
=(lg5)^2 + lg2*(lg5+1)
=(lg5+lg2)*lg5 + lg2
=lg5+lg2
=1
(公式lgx + lgy=lg(xy))

=lg(5^5)+lg(2^50)
=lg(5^5*2^50)
=lg(10^5*2^45)
=45lg2

(lg5)^2 + lg2*lg50
=(1-lg2)^2+lg2*(lg100-lg2)-----(lg2=lg(10/2)=lg10-lg2=1-lg2)
=1 ------(展开)

lg50=1g(2*5*5)=lg2+2lg5
原式=(lg5)^2+(lg2)^2+2lg2lg5
=（lg5+lg2)^2=(lg10)^2=1