Question

高一数学题！！！！！！！！！！

1、lg5^2+2/3lg8+lg5lg20+(lg2)^2
2、（log2^5+log4^0.2)(log5^2+log25^0.5)
3、若lg(x-y)+lg(x+2y)=lg2+lgx+lgy，求x/y的值.
求解答！！追加悬赏分！！！！

Answer 1

你好！

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

如有疑问可追问。祝你开心！

Answer 2

文化低 飘过