\begin{array}{l}已知函数f(x)=x-\frac{1}{x+1},g(x)={x}^{2}-2ax+4,若∀{x}_{1}∈[0,1],∃{x}_{2}∈[1,2],\\ 使f({x}_{1})≥g({x}_{2}),则实数a的取值范围是\end{array}

\begin{array}{l}已知函数f(x)=x-\frac{1}{x+1},g(x)={x}^{2}-2ax+4,若∀{x}_{1}∈[0,1],∃{x}_{2}∈[1,2],\\ 使f({x}_{1})≥g({x}_{2}),则实数a的取值范围是\end{array}
\begin{array}{l}已知函数f(x)=x-\frac{1}{x+1},g(x)={x}^{2}-2ax+4,若∀{x}_{1}∈[0,1],∃{x}_{2}∈[1,2],\\ 使f({x}_{1})≥g({x}_{2}),则实数a的取值范围是\end{array}

\begin{array}{l}已知函数f(x)=x-\frac{1}{x+1},g(x)={x}^{2}-2ax+4,若∀{x}_{1}∈[0,1],∃{x}_{2}∈[1,2],\\ 使f({x}_{1})≥g({x}_{2}),则实数a的取值范围是\end{array}
f(x)=x-1/(1+x)在[0,1]单调增加 其最小值为f(0)=-1
故g(x)=x^2-2ax+4≤-1 在[1,2]恒成立 令x=1 可得到 a≥3>2 故g(x)在[1,2]单调减小 只需要g(1)≤-1
解得 a≥3