如果方程loga(x-5)/(x+5)=1+loga(x-3)有实根 求实数a的取值

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就当真数是(x-5)/(x+5)
loga(x-5)/(x+5)=1+loga(x-3)
loga(x-5)/(x+5)=loga[a(x-3)]
(x-5)/(x+5)=a(x-3)
x-5=a(x-3)(x+5)
ax^2+(2a-1)x+(5-15a)=0
要使方程有实根,则Δ≥0
(2a-1)^2-4a(5-15a)≥0
64a^2-24a+1≥0
0======以下答案可供参考======
供参考答案1:
这个要分类讨论了~