288. $y=\frac{3 x^{2}-4 x+1}{2 x+1}$.

288. $y=\frac{3 x^{2}-4 x+1}{2 x+1}$.
Solution. We apply the quotient rule of differentiation and formula $\mathrm{X}$ :

$$
\begin{gathered}
y^{\prime}=\frac{\left(3 x^{2}-4 x+1\right)^{\prime}(2 x+1)-(2 x+1)^{\prime}\left(3 x^{2}-4 x+1\right)}{(2 x+1)^{2}}= \\
=\frac{(6 x-4)(2 x+1)-2\left(3 x^{2}-4 x+1\right)}{(2 x+1)^{2}}=\frac{12 x^{2}+6 x-8 x-4-6 x^{2}+8 x-2}{(2 x+1)^{2}}= \\
=\frac{6 x^{2}+6 x-6}{(2 x+1)^{2}}=\frac{6\left(x^{2}+x-1\right)}{(2 x+1)^{2}}
\end{gathered}
$$