Question

若lg(x-y)+lg(x+2y)=lg2+lgx+lgy,求x/y的值？

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） 上面都懂 ， 从设开始就不懂了，为什么a^2-a-2 =0 ？

Answer 1

x/y - 2y/x = 1(等式两同时乘以x/y）
x/y *x/y - 2y/x*x/y =x/y
(x/y)^2 - 2 =x/y
(x/y)^2-x/y - 2 =0
a = x/y
则a^2 - a - 2 = 0
(a - 2 )(a + 1) = 0

Answer 2

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

lg2 + 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)^2 - 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
( lgx + lgy=lgxy)对数的性质

Answer 3

设之后变为a-2*1/a=1两边同乘以a再移项就得到了