将 $$$u^{6}$$$ 除以 $$$u^{2} + 1$$$
您的输入
使用长除法计算$$$\frac{u^{6}}{u^{2} + 1}$$$。
解答
将题目写成特殊格式(缺失项写为零系数):
$$$\begin{array}{r|r}\hline\\u^{2}+1&u^{6}+0 u^{5}+0 u^{4}+0 u^{3}+0 u^{2}+0 u+0\end{array}$$$
步骤 1
将被除式的首项除以除式的首项: $$$\frac{u^{6}}{u^{2}} = u^{4}$$$.
将计算结果写在表格的上部。
将其乘以除数:$$$u^{4} \left(u^{2}+1\right) = u^{6}+u^{4}$$$。
从得到的结果中减去被除数:$$$\left(u^{6}\right) - \left(u^{6}+u^{4}\right) = - u^{4}$$$
$$\begin{array}{r|rrrrrrr:c}&{\color{Red}u^{4}}&&&&&&&\\\hline\\{\color{Magenta}u^{2}}+1&{\color{Red}u^{6}}&+0 u^{5}&+0 u^{4}&+0 u^{3}&+0 u^{2}&+0 u&+0&\frac{{\color{Red}u^{6}}}{{\color{Magenta}u^{2}}} = {\color{Red}u^{4}}\\&-\phantom{u^{6}}&&&&&&&\\&u^{6}&+0 u^{5}&+u^{4}&&&&&{\color{Red}u^{4}} \left(u^{2}+1\right) = u^{6}+u^{4}\\\hline\\&&&- u^{4}&+0 u^{3}&+0 u^{2}&+0 u&+0&\end{array}$$步骤 2
将所得余式的首项除以除式的首项: $$$\frac{- u^{4}}{u^{2}} = - u^{2}$$$
将计算结果写在表格的上部。
将其乘以除数:$$$- u^{2} \left(u^{2}+1\right) = - u^{4}- u^{2}$$$。
从得到的结果中减去余数:$$$\left(- u^{4}\right) - \left(- u^{4}- u^{2}\right) = u^{2}$$$
$$\begin{array}{r|rrrrrrr:c}&u^{4}&{\color{Peru}- u^{2}}&&&&&&\\\hline\\{\color{Magenta}u^{2}}+1&u^{6}&+0 u^{5}&+0 u^{4}&+0 u^{3}&+0 u^{2}&+0 u&+0&\\&-\phantom{u^{6}}&&&&&&&\\&u^{6}&+0 u^{5}&+u^{4}&&&&&\\\hline\\&&&{\color{Peru}- u^{4}}&+0 u^{3}&+0 u^{2}&+0 u&+0&\frac{{\color{Peru}- u^{4}}}{{\color{Magenta}u^{2}}} = {\color{Peru}- u^{2}}\\&&&-\phantom{- u^{4}}&&&&&\\&&&- u^{4}&+0 u^{3}&- u^{2}&&&{\color{Peru}- u^{2}} \left(u^{2}+1\right) = - u^{4}- u^{2}\\\hline\\&&&&&u^{2}&+0 u&+0&\end{array}$$步骤 3
将所得余式的首项除以除式的首项: $$$\frac{u^{2}}{u^{2}} = 1$$$
将计算结果写在表格的上部。
将其乘以除数:$$$1 \left(u^{2}+1\right) = u^{2}+1$$$。
从得到的结果中减去余数:$$$\left(u^{2}\right) - \left(u^{2}+1\right) = -1$$$
$$\begin{array}{r|rrrrrrr:c}&u^{4}&- u^{2}&{\color{Chartreuse}+1}&&&&&\\\hline\\{\color{Magenta}u^{2}}+1&u^{6}&+0 u^{5}&+0 u^{4}&+0 u^{3}&+0 u^{2}&+0 u&+0&\\&-\phantom{u^{6}}&&&&&&&\\&u^{6}&+0 u^{5}&+u^{4}&&&&&\\\hline\\&&&- u^{4}&+0 u^{3}&+0 u^{2}&+0 u&+0&\\&&&-\phantom{- u^{4}}&&&&&\\&&&- u^{4}&+0 u^{3}&- u^{2}&&&\\\hline\\&&&&&{\color{Chartreuse}u^{2}}&+0 u&+0&\frac{{\color{Chartreuse}u^{2}}}{{\color{Magenta}u^{2}}} = {\color{Chartreuse}1}\\&&&&&-\phantom{u^{2}}&&&\\&&&&&u^{2}&+0 u&+1&{\color{Chartreuse}1} \left(u^{2}+1\right) = u^{2}+1\\\hline\\&&&&&&&-1&\end{array}$$由于余式的次数小于除式的次数,故除法完成。
所得表格再次显示如下:
$$\begin{array}{r|rrrrrrr:c}&{\color{Red}u^{4}}&{\color{Peru}- u^{2}}&{\color{Chartreuse}+1}&&&&&\text{提示}\\\hline\\{\color{Magenta}u^{2}}+1&{\color{Red}u^{6}}&+0 u^{5}&+0 u^{4}&+0 u^{3}&+0 u^{2}&+0 u&+0&\frac{{\color{Red}u^{6}}}{{\color{Magenta}u^{2}}} = {\color{Red}u^{4}}\\&-\phantom{u^{6}}&&&&&&&\\&u^{6}&+0 u^{5}&+u^{4}&&&&&{\color{Red}u^{4}} \left(u^{2}+1\right) = u^{6}+u^{4}\\\hline\\&&&{\color{Peru}- u^{4}}&+0 u^{3}&+0 u^{2}&+0 u&+0&\frac{{\color{Peru}- u^{4}}}{{\color{Magenta}u^{2}}} = {\color{Peru}- u^{2}}\\&&&-\phantom{- u^{4}}&&&&&\\&&&- u^{4}&+0 u^{3}&- u^{2}&&&{\color{Peru}- u^{2}} \left(u^{2}+1\right) = - u^{4}- u^{2}\\\hline\\&&&&&{\color{Chartreuse}u^{2}}&+0 u&+0&\frac{{\color{Chartreuse}u^{2}}}{{\color{Magenta}u^{2}}} = {\color{Chartreuse}1}\\&&&&&-\phantom{u^{2}}&&&\\&&&&&u^{2}&+0 u&+1&{\color{Chartreuse}1} \left(u^{2}+1\right) = u^{2}+1\\\hline\\&&&&&&&-1&\end{array}$$因此,$$$\frac{u^{6}}{u^{2} + 1} = \left(u^{4} - u^{2} + 1\right) + \frac{-1}{u^{2} + 1}$$$。
答案
$$$\frac{u^{6}}{u^{2} + 1} = \left(u^{4} - u^{2} + 1\right) + \frac{-1}{u^{2} + 1}$$$A