Question

一道高一数学题，解决暑假作业嘛

若lg(x-y)+lg(x+2y)=lg2+lgx+lgy,求x/y的值

Answer 1

lg(x - y) + lg(x + 2y)
= lg[(x - y)(x + 2y)]

lg^2 + lgx + lgy
= lg(2xy)

所以：
(x - y)(x + 2y) = 2xy
x^2 + 2xy - xy - 2y^2 = 2xy
x^2 - 2y^2 = xy
x/y - 2y/x = 1

设 a = x/y 则有

a^2 - a - 2 = 0
(a - 2 )(a + 1) = 0
a = 2,-1 ..............根据题意取正
所以：x/y = 2

Answer 2

lg(x-y)+lg(x+2y)=lg2+lgx+lgy
lg(x-y)(x+2y)=lg2xy
(x-y)(x+2y)=2xy
x^2+2xy-xy-2y^2=2xy
x^2-xy-2y^2=0
(x-2y)(x+y)=0
x=2y 或 x+y=0
因为x>0 y>0
所以x+y=0(舍去）
所以x/y=2