Đáp án:
\(\begin{array}{l}
\dfrac{2}{2+x^2} = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{4} - \dfrac{x^6}{8} + o(x^7)
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
\quad f(x) = \dfrac{2}{2+x^2}\\
+)\quad f(0) = 1\\
+)\quad f'(x) = - \dfrac{4x}{(2+x^2)^2} \Rightarrow f'(0) = 0\\
+)\quad f''(x) = \dfrac{4(3x^2 - 2)}{(2+x^2)^3} \Rightarrow f''(0) = -1\\
+)\quad f'''(x) = -\dfrac{48x(x^2 - 2)}{(2+x^2)^4} \Rightarrow f'''(0) = 0\\
+)\quad f^{(4)}(x) = \dfrac{48(5x^4 - 20x^2 + 4)}{(2+x^2)^5} \Rightarrow f^{(4)}(0) = 6\\
+)\quad f^{(5)}(x) = -\dfrac{480x(3x^4 - 20x^2 + 12)}{(2+x^2)^6} \Rightarrow f^{(5)}(0) = 6\\
+)\quad f^{(6)}(x) = \dfrac{1440(7x^6-70x^4 + 84x^2 -8)}{(2+x^2)^7} \Rightarrow f^{(6)}(0) = -90\\
\text{Ta được:}\\
f(x) = f(0) + \dfrac{f'(0)}{1!}x + \dfrac{f''(0)}{2!}x^2 + \dfrac{f'''(0)}{3!}x^3 + \dfrac{f^{(4)}(0)}{4!}x^4 + \dfrac{f^{(5)}(0)}{5!}x^5 + \dfrac{f^{(6)}(0)}{6!}x^6 + o(x^7)\\
\Leftrightarrow f(x) = 1 - \dfrac{x^2}{2} + \dfrac{x^4}{4} - \dfrac{x^6}{8} + o(x^7)
\end{array}\)
