.  - 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