Question

方程lgx2-lg(x+3)=lga(a∈(0,+∞))在区间(3,4)内有解,则a的取值范围为 .

Answer 1

不用导数照样简单化简为 lg(x²/(x+3) )=lga∴x²/(x+3)=a令t=x+3,t∈(6,7),x=t-3,则f(x)=x²/(x+3)可化简为f(t)=(t-3)²/t=t²-6t+9 / t=t+(9/t)-6而对勾函数t+(9/t)在(6,7)为增,则f(t)=t+(9/t...