15．已知 $f(x)=(x+a+b)\ln\left(1+\dfrac{1}{x+b}\right)$，若 $f(x)$ 为偶函数，则 $a=$ __．
解答
$$f(x)=(x+a+b)\ln\left(1+\frac{1}{x+b}\right)=(x+a+b)\ln\frac{x+b+1}{x+b}$$
$$f(-x)=(-x+a+b)\ln\left(1+\frac{1}{-x+b}\right)=(-x+a+b)\ln\frac{x-b-1}{x-b}$$
$\because f(-x)=f(x)$
$\therefore$ 考虑 $a+b=0$，$f(-x)=-x\ln\dfrac{x-b-1}{x-b}=x\ln\dfrac{x-b}{x-b-1}$
$$\therefore\ \frac{x-b}{x-b-1}=\frac{x+b+1}{x+b}$$
$$\therefore\ x^2-b^2=x^2-(b+1)^2$$
$$\therefore\ -b=b+1$$
$$\therefore\ a=-b=\frac{1}{2}$$
答案： $a=\dfrac{1}{2}$