$\begin{array}{l}
y = \frac{{2x + 1}}{{1 - x}} \Rightarrow y' = \frac{3}{{{{\left( {1 - x} \right)}^2}}}\\
\left( C \right)\,cat\,Oy\,tai\,\left( {0;1} \right)\,nen\,tiep\,tuyen\,co\,phuong\,trinh:\\
y = y'\left( 0 \right)\left( {x - 0} \right) + 1 = 3x + 1
\end{array}$