a/ $y=\frac{2 x+3}{x^{2}+x+3}$

Q: a/ $y=\frac{2 x+3}{x^{2}+x+3}$

$\mathrm{a} / y=\frac{2 x+3}{x^{2}+x+3} \Rightarrow \mathrm{y}^{\prime}=\frac{(2 \mathrm{x}+3)^{\prime}\left(\mathrm{x}^{2}+\mathrm{x}+3\right)-(2 \mathrm{x}+3)\left(\mathrm{x}^{2}+\mathrm{x}+3\right)^{\prime}}{\left(\mathrm{x}^{2}+\mathrm{x}+3\right)}$ $=\frac{2\left(x^{2}+x+3\right)-(2 x+3)(2 x+1)}{\left(x^{2}+x+3\right)^{2}}=\frac{-2 x^{2}-6 x+3}{\left(x^{2}+x+3\right)^{2}}$

Question

Accepted Answer

$\mathrm{a} / y=\frac{2 x+3}{x^{2}+x+3} \Rightarrow \mathrm{y}^{\prime}=\frac{(2 \mathrm{x}+3)^{\prime}\left(\mathrm{x}^{2}+\mathrm{x}+3\right)-(2 \mathrm{x}+3)\left(\mathrm{x}^{2}+\mathrm{x}+3\right)^{\prime}}{\left(\mathrm{x}^{2}+\mathrm{x}+3\right)}$

$=\frac{2\left(x^{2}+x+3\right)-(2 x+3)(2 x+1)}{\left(x^{2}+x+3\right)^{2}}=\frac{-2 x^{2}-6 x+3}{\left(x^{2}+x+3\right)^{2}}$

THPT Phan Ngọc Hiển - Đề thi cuối kì 2 (CT) 20-21 - Cà Mau - MĐ 6652