想必現(xiàn)在有很多小伙伴對于判斷并改正$\dfrac{{\displaystyle \dfrac{{\displaystyle 6\underset{\begin{array}{lll}6& & \end{array}}{\overset{\begin{array}{lll}& 1& 6\end{array}}{\require{enclose}\enclose{longdiv}{ \begin{array}{lll}6& 3& 6\end{array}}}}}}{\begin{array}{llll}& & 3& 6\\ & & 3& 6\end{array}}}}{\begin{array}{llll}& & & 0\end{array}}$（）改正：","title_text":"判斷并改正$\dfrac{{\displaystyle \dfrac{{\displaystyle 6\underset{\begin{array}{lll}6 \end{array}}{\overset{\begin{array}{lll} 1 6\end{array}}{\require{enclose}\enclose{longdiv}{ \begin{array}{lll}6 3 6\end{array}}}}}}{\begin{array}{llll} 3 6\\ 3 6\end{array}}}}{\begin{array}{llll} 0\end{array}}$（）改正：方面的知識都比較想要了解，那么今天小好小編就為大家收集了一些關于判斷并改正$\dfrac{{\displaystyle \dfrac{{\displaystyle 6\underset{\begin{array}{lll}6& & \end{array}}{\overset{\begin{array}{lll}& 1& 6\end{array}}{\require{enclose}\enclose{longdiv}{ \begin{array}{lll}6& 3& 6\end{array}}}}}}{\begin{array}{llll}& & 3& 6\\ & & 3& 6\end{array}}}}{\begin{array}{llll}& & & 0\end{array}}$（）改正：","title_text":"判斷并改正$\dfrac{{\displaystyle \dfrac{{\displaystyle 6\underset{\begin{array}{lll}6 \end{array}}{\overset{\begin{array}{lll} 1 6\end{array}}{\require{enclose}\enclose{longdiv}{ \begin{array}{lll}6 3 6\end{array}}}}}}{\begin{array}{llll} 3 6\\ 3 6\end{array}}}}{\begin{array}{llll} 0\end{array}}$（）改正：方面的知識分享給大家，希望大家會喜歡哦。

【答案】

?

改正：

【解析】

改正：

故答案為：?

本文到此結束，希望對大家有所幫助。

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