\(\exists x\geqslant 0,f(x)=x\ln(x+1)+(\dfrac{1}{2}-a)x+2-a<0\)
\(\Rightarrow \exists x\geqslant 0,x\ln(x+1)+\dfrac{1}{2}x+2<a(x+1)\)
令\(g(x)=x\ln(x+1)+\dfrac{1}{2}x+2\)，则
\(g'(x)=\ln(x+1)+\dfrac{x}{x+1}+\dfrac{1}{2}=\ln(x+1)-\dfrac{1}{x+1}+\dfrac{3}{2}\)
函数\(g(x)\)在点\(x=x_0\)处的切线的斜率为\(\ln(x_0+1)-\dfrac{1}{x_0+1}+\dfrac{3}{2}\)
\(\Rightarrow\)函数\(g(x)\)在点\(x=x_0\)处的切线方程\(y-(x_0\ln(x_0+1)+\dfrac{1}{2}x_0+2)=(\ln(x_0+1)-\dfrac{1}{x_0+1}+\dfrac{3}{2})(x-x_0)\)与直线\(y=a(x+1)\)重合， \(\Rightarrow \left\{ \begin{array}{ll} \ln(x_0+1)+\dfrac{x_0}{x_0+1}+\frac{1}{2}=a \\ x_0\ln(x_0+1)+\frac{1}{2}x_0+2=a(x_0+1) \end{array} \right. \Rightarrow \left\{ \begin{array}{ll} a=4-(x_0+1)-\frac{1}{x_0+1} \\ \ln(x_0+1)+x_0-\frac{3}{2}=0\Rightarrow 0<x_0<1 \end{array} \right.\)
\(\Rightarrow \dfrac{3}{2}<4-(x_0+1)-\frac{1}{x_0+1} <2\)
综上可知，\(a=2.\)(图还未配)
\(f'(x)=\ln(x+1)+\dfrac{x}{x+1}+\dfrac{1}{2}-a=\ln(x+1)-\dfrac{1}{x+1}+\dfrac{3}{2}-a\)
\(\Rightarrow f'(x)\)单调递增\(\Rightarrow f'(0)=\dfrac{1}{2}-a\)
\(1^o\)当\(a\leqslant \dfrac{1}{2}\)时，\(f'(x)\geqslant f'(0)=\dfrac{1}{2}-a\geqslant 0\Rightarrow f(x)_{min}=f(0)=2-a>0\)，不合题意；
\(2^o\)当\(a=1\)时，\(f'(x)=\ln(x+1)-\dfrac{1}{x+1}+\dfrac{1}{2}\Rightarrow f'(0)=-\dfrac{1}{2}<0,f'(1)=\ln 2>0\)
\(\Rightarrow f'(x_0)=0\Rightarrow 0<x_0<1\Rightarrow \left\{ \begin{array}{ll} \ln(x_0+1)-\dfrac{1}{x_0+1}+\dfrac{1}{2}=0 \\ f(x_0)=x_0\ln(x_0+1)-\frac{1}{2}x_0+\dfrac{3}{2} \end{array} \right.\)
\(\Rightarrow f(x_0)=-\frac{1}{x_0+1}-(x_0+1)+\dfrac{7}{2}\in (1,\dfrac{3}{2})\)，不合题意；
\(3^o\)当\(a=2\)时，\(f(\frac{1}{2})=\dfrac{1}{2}\ln\dfrac{3}{2}-\dfrac{3}{4}<0\)，合题意\(.\)
综上可知，\(a=2.\)
