割り算の等式 A=BQ+R ➡ Bを求めよう

次の条件を満たす整式 B を求めよう。

この問題へのリンクはこちら(右クリックで保存)

\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
\colBX{bisque}{$A$}\ を &\ \colBX{palegreen}{$B$}\ で割ると,商が\ \colBX{violet}{$Q$},余りが\ \colBX{lightcyan}{$R$}\\
\\
& \colNS{red}{\Darr\ 4つの要素がそろった!}\\
\\
\Large \colBX{bisque}{$A$} & \Large = \colBX{palegreen}{$B$} \times \colBX{violet}{$Q$} + \colBX{lightcyan}{$R$}
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】

\def\A{x^3+2x-1}
\def\Q{x+2}
\def\R{6x-1}
\def\Tenkai{x^3+2x-1-6x+1}
\def\Seiri{x^3-4x}
\def\Kotae{x^2-2x}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
整式\ \colBX{bisque}{$\A$}\ を\ \colBX{palegreen}{$B$}\ & で割ると,\\
商が\ \colBX{violet}{$\Q$},余りが & \ \colBX{lightcyan}{$\R$}\ であるから\\
\\
\colBX{bisque}{$\A$} = & \colBX{palegreen}{$B$} \times (\colBX{violet}{$\Q$})+\colBX{lightcyan}{$\R$}\\
& \colMM{red}{\swarrow 余りを移項}\\
\A -(\R)= &\ B \times (\Q)\\
\colMM{red}{左右\ \searrow\ } & \colMM{red}{\swarrow 入れ替え}\\
B \times (\Q) = &\ \Tenkai\\
\\
B \times (\Q) = &\ \Seiri\\
\colMM{red}{両辺を商\ } & \colMM{red}{(\Q)\ で割る}\\
B =&\ (\Seiri) \div (\Q)\\
& \colMM{red}{下の計算より}\\
=&\ \Kotae
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【割り算】

\def\B{x+2}
\def\Qa{}
\def\Qb{-2x}
\def\Qc{x^2}
\def\Qd{}

\def\Aa{}
\def\Ab{-4x}
\def\Ac{}
\def\Ad{x^3}

\def\Ba{}
\def\Bb{}
\def\Bc{+2x^2}
\def\Bd{x^3}

\def\Ca{}
\def\Cb{-4x}
\def\Cc{-2x^2}
\def\Cd{}

\def\Da{}
\def\Db{-4x}
\def\Dc{-2x^2}
\def\Dd{}

\def\Ea{}
\def\Eb{0}
\def\Ec{}
\def\Ed{}



\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{array}{rrrrrrl}
& & \Qd  & \Qc & \Qb & \Qa & \colMM{violet}{\Leftarrow 商}\\\hline
\colBX{palegreen}{$\B$} & \textcolor{red}{\big)} & \Ad  & \Ac & \Ab & \Aa & \\
& & \Bd  & \Bc & \Bb & \Ba & \colMM{orange}{\Leftarrow (\B) \times \Qc}\\\hline
& & \Cd  & \Cc & \Cb & \Ca \\
& & \Dd  & \Dc & \Db & \Da &  \colMM{orange}{\Leftarrow (\B) \times (\Qb)}\\\hline
& & \Ed  & \Ec & \Eb & \Ea & \colBX{lightcyan}{\scriptsize\bf\color{deepskyblue} 余りは0になる!}\\
\end{array}%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

この問題へのリンクはこちら(右クリックで保存)

\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
\colBX{bisque}{$A$}\ を &\ \colBX{palegreen}{$B$}\ で割ると,商が\ \colBX{violet}{$Q$},余りが\ \colBX{lightcyan}{$R$}\\
\\
& \colNS{red}{\Darr\ 4つの要素がそろった!}\\
\\
\Large \colBX{bisque}{$A$} & \Large = \colBX{palegreen}{$B$} \times \colBX{violet}{$Q$} + \colBX{lightcyan}{$R$}
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】

\def\A{3x^2-4x+5}
\def\Q{x-1}
\def\R{4}
\def\Tenkai{\A -4}
\def\Seiri{3x^2-4x+1}
\def\Kotae{3x-1}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
整式\ \colBX{bisque}{$\A$}\ を\ \colBX{palegreen}{$B$}\ & で割ると,\\
商が\ \colBX{violet}{$\Q$},余りが & \ \colBX{lightcyan}{$\R$}\ であるから\\
\\
\colBX{bisque}{$\A$} = & \colBX{palegreen}{$B$} \times (\colBX{violet}{$\Q$})+\colBX{lightcyan}{$\R$}\\
& \colMM{red}{\swarrow 余りを移項}\\
\A -\R= &\ B \times (\Q)\\
\colMM{red}{左右\ \searrow\ } & \colMM{red}{\swarrow 入れ替え}\\
B \times (\Q) = &\ \Tenkai\\
\\
B \times (\Q) = &\ \Seiri\\
\colMM{red}{両辺を商\ } & \colMM{red}{(\Q)\ で割る}\\
B =&\ (\Seiri) \div (\Q)\\
& \colMM{red}{下の計算より}\\
=&\ \Kotae
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【割り算】

\def\B{x-1}
\def\Qa{-1}
\def\Qb{3x}
\def\Qc{}
\def\Qd{}

\def\Aa{+1}
\def\Ab{-4x}
\def\Ac{3x^2}
\def\Ad{}

\def\Ba{}
\def\Bb{-3x}
\def\Bc{3x^2}
\def\Bd{}

\def\Ca{+1}
\def\Cb{-x}
\def\Cc{}
\def\Cd{}

\def\Da{+1}
\def\Db{-x}
\def\Dc{}
\def\Dd{}

\def\Ea{0}
\def\Eb{}
\def\Ec{}
\def\Ed{}



\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{array}{rrrrrrl}
& & \Qd  & \Qc & \Qb & \Qa & \colMM{violet}{\Leftarrow 商}\\\hline
\colBX{palegreen}{$\B$} & \textcolor{red}{\big)} & \Ad  & \Ac & \Ab & \Aa & \\
& & \Bd  & \Bc & \Bb & \Ba & \colMM{orange}{\Leftarrow (\B) \times \Qb}\\\hline
& & \Cd  & \Cc & \Cb & \Ca \\
& & \Dd  & \Dc & \Db & \Da &  \colMM{orange}{\Leftarrow (\B) \times (\Qa)}\\\hline
& & \Ed  & \Ec & \Eb & \Ea & \colBX{lightcyan}{\scriptsize\bf\color{deepskyblue} 余りは0になる!}\\
\end{array}%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

この問題へのリンクはこちら(右クリックで保存)

\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\textcolor{#1}{\scriptsize\bf\bm #2}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
\colBX{bisque}{$A$}\ を &\ \colBX{palegreen}{$B$}\ で割ると,商が\ \colBX{violet}{$Q$},余りが\ \colBX{lightcyan}{$R$}\\
\\
& \colNS{red}{\Darr\ 4つの要素がそろった!}\\
\\
\Large \colBX{bisque}{$A$} & \Large = \colBX{palegreen}{$B$} \times \colBX{violet}{$Q$} + \colBX{lightcyan}{$R$}
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【解答】

\def\A{x^3-2x^2+3x-3}
\def\Q{x-2}
\def\R{-2x+7}
\def\Tenkai{\A +2x-7}
\def\Seiri{x^3-2x^2+5x-10}
\def\Kotae{x^2+5}
\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{align*}
整式\ \colBX{bisque}{$\A$}\ を\ \colBX{palegreen}{$B$}\ & で割ると,\\
商が\ \colBX{violet}{$\Q$},余りが & \ \colBX{lightcyan}{$\R$}\ であるから\\
\\
\colBX{bisque}{$\A$} = & \colBX{palegreen}{$B$} \times (\colBX{violet}{$\Q$})\colBX{lightcyan}{$\R$}\\
& \colMM{red}{\swarrow 余りを移項}\\
\A -(\R)= &\ B \times (\Q)\\
\colMM{red}{左右\ \searrow\ } & \colMM{red}{\swarrow 入れ替え}\\
B \times (\Q) = &\ \Tenkai\\
\\
B \times (\Q) = &\ \Seiri\\
\colMM{red}{両辺を商\ } & \colMM{red}{(\Q)\ で割る}\\
B =&\ (\Seiri) \div (\Q)\\
& \colMM{red}{下の計算より}\\
=&\ \Kotae
\end{align*}
%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

【割り算】

\def\B{x-2}
\def\Qa{+5}
\def\Qb{}
\def\Qc{x^2}
\def\Qd{}

\def\Aa{-10}
\def\Ab{+5x}
\def\Ac{-2x^2}
\def\Ad{x^3}

\def\Ba{}
\def\Bb{}
\def\Bc{-2x^2}
\def\Bd{x^3}

\def\Ca{-10}
\def\Cb{5x}
\def\Cc{}
\def\Cd{}

\def\Da{-10}
\def\Db{5x}
\def\Dc{}
\def\Dd{}

\def\Ea{0}
\def\Eb{}
\def\Ec{}
\def\Ed{}



\newcommand\colNS[2]{\color{#1}#2\color{black}}
\newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}}
\newcommand\colMM[2]{\color{#1}\scriptsize #2\color{black}}
\newcommand\colBX[2]{\colorbox{#1}{#2}}
\newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}}
\begin{array}{rrrrrrl}
& & \Qd  & \Qc & \Qb & \Qa & \colMM{violet}{\Leftarrow 商}\\\hline
\colBX{palegreen}{$\B$} & \textcolor{red}{\big)} & \Ad  & \Ac & \Ab & \Aa & \\
& & \Bd  & \Bc & \Bb & \Ba & \colMM{orange}{\Leftarrow (\B) \times \Qc}\\\hline
& & \Cd  & \Cc & \Cb & \Ca \\
& & \Dd  & \Dc & \Db & \Da &  \colMM{orange}{\Leftarrow (\B) \times (\Qa)}\\\hline
& & \Ed  & \Ec & \Eb & \Ea & \colBX{lightcyan}{\scriptsize\bf\color{deepskyblue} 余りは0になる!}\\
\end{array}%1 orange,bisque
%2 green,palegreen
%3 magenta, violet
%4 deepskyblue, lightcyan

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