Tính f (x) = căn (2x + 3) khai triển taylor đến cấp 3 tại x0 = 3

$\begin{array}{l}\begin{array}{|l|cr|}
\hline
f(x) = \sqrt{2x + 3}&f(3) = 3\\
\hline
f'(x) = \dfrac{1}{\sqrt{2x + 3}}&f'(3) = \dfrac13\\
\hline
f''(x) = -\dfrac{1}{\sqrt{(2x+3)^3}}&f''(3) = - \dfrac{1}{27}\\
\hline
f'''(x) = \dfrac{3}{\sqrt{(2x + 3)^5}}&f'''(3) = \dfrac{1}{81}\\
\hline
f^{(4)}(x) = -\dfrac{15}{\sqrt{(2x +3)^7}}&f^{(4)}(c) =-\dfrac{15}{\sqrt{(2c +3)^7}}\\
\hline
\end{array} \\
\text{Ta được:}\\
f(x) = f(1) + \dfrac{f'(3)}{1!}(x-3) + \dfrac{f''(3)}{2!}(x-3)^2 + \dfrac{f'''(3)}{3!}(x-3)^3 +\dfrac{f^{(4)}(c)}{4!}(x-3)^4\\
\to f(x) = 3 + \dfrac{1}{3}(x-3)-\dfrac{1}{54}(x-3)^2 +\dfrac{1}{486}(x-3)^3 -\dfrac{5}{8\sqrt{(2c +3)^7}}(x-3)^4
\end{array}$