1、原式= 2lg5 +(2/3)lg2³ +lg5 * lg(2²×5) + (lg2)²
= 2lg5 + 2lg2 + (lg5) (2lg2 +lg5) +(lg2)²
= 2lg5 +2lg2 +2(lg2)(lg5) +(lg5)² +(lg2)²
= 2(lg5 +lg2) +(lg5 + lg2)²
=2lg10 +(lg10)²
= 3
2、原式= [(lg5)/(lg2) + (lg0.2)/(lg4)] [(lg2)/(lg5) + (lg0.5)/(lg25)]
= [(lg5)/(lg2) - (lg5)/(2lg2)] [(lg2)/(lg5) - (lg2)/(2lg5)]
= (lg5)/(2lg2) * (lg2)/(2lg5)
=1/4
3、lg(x-y)+lg(x+2y) = lg[(x-y)(x+2y)]
lg2+lgx+lgy =lg(2xy)
∴(x-y)(x+2y) = 2xy
x²+xy-2y²=2xy
x²-xy-2y²=0
两边同除以y²
(x/y)² - x/y -2 =0
解出x/y = -1或2
因为x,y>0,所以舍去负值
故x/y =2