lg(-x²+3x+m)=lg(3-x)
-x²+3x+m = 3 - x
x² - 4x - m = 0
∆ = 16 + 4m = 4(4 + m)
(1) ∆ < 0
m < -4:无解
(2) ∆ = 0
m = -4
此时 -x²+3x+m = -x²+3x-4 = -(x - 3/2)² - 7/4 < 0
无解
(3) ∆ > 0
m > -4
f(x) = -x²+3x+m = -(x- 3/2)² + m + 9/4
(i)m < -9/4时,f(x)恒为负,无解
(ii) m = -9/4时,f(x)最大值为0,ln[f(x)]无意义,无解
(iii) m > -9/4
x² - 4x - m = 0的解为:x₁ = 2 + √(4+ m),x₂ = 2 - √(4+m)
(a) 0 < x₁ < 3
0 < 2 + √(4+ m) < 3
-2 < √(4+ m) < 1
只需考虑右边一半:4 + m < 1,m < -3,与前提m> -9/4矛盾,舍去
(b) 0 < x₂ < 3
0 < 2 - √(4+m) < 3
-2 < -√(4+m) < 1
2 > √(4+m) > -1
只需考虑左边一半:2 > √(4+m),4 > 4 + m,m < 0
结合前提:-9/4 < m < 0
即 -9/4 < m < 0时,方程有一解; m取其它值时,方程无解