Câu hỏi 29

 Ta có \(f\left( x \right)=\frac{{{x}^{2}}}{1-x}=\frac{{{x}^{2}}-1+1}{1-x}=-\,1-x+\frac{1}{1-x}\Rightarrow \,\,{f}'\left( x \right)=-\,1+\frac{1}{{{\left( 1-x \right)}^{2}}}=-\,1+{{\left( 1-x \right)}^{-\,2}}.\)

Suy ra \({f}''\left( x \right)=\,2.{{\left( 1-x \right)}^{-\,3}}=2!.{{\left( 1-x \right)}^{-\,3}}\Rightarrow \,\,{f}'''\left( x \right)=2.3.{{\left( 1-x \right)}^{-\,4}}=3!.{{\left( 1-x \right)}^{-\,4}}.\)

Vậy \({{f}^{\left( 30 \right)}}\left( x \right)=30!.{{\left( 1-x \right)}^{-\,31}}.\) Chọn B.