次の式を計算せよ。
↓この問題へのリンクはこちら(右クリックで保存)
【解答】
\def\BunsiA{2} \def\BunboA{x+1} \def\BunsiB{1} \def\BunboB{x-3} \def\Tenkai{2x-6+x+1} \def\Kotae{3x-5} \newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize\bf\bm #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} &\dfrac{\BunsiA}{\colBX{mistyrose}{$\BunboA$}} + \dfrac{\BunsiB}{\colBX{palegreen}{$\BunboB$}}\\ & \colMM{green}{相手の分母からもらう \nwarrow} \colMM{orange}{\nearrow 相手の分母からもらう}\\ &= \dfrac{\BunsiA \colBX{palegreen}{$(\BunboB$)}}{\colBX{mistyrose}{$(\BunboA)$}\colBX{palegreen}{$(\BunboB)$}} + \dfrac{\BunsiB\colBX{mistyrose}{$(\BunboA)$}}{\colBX{palegreen}{$(\BunboB)$}\colBX{mistyrose}{$(\BunboA)$}}\\ & \colMM{red}{分母が同じ➡分子だけ計算する}\\ &= \dfrac{\BunsiA(\BunboB)+\BunsiB(\BunboA)}{(\BunboA)(\BunboB)}\\ \\ &= \dfrac{\Tenkai}{(\BunboA)(\BunboB)}\\ \\ &= \dfrac{\Kotae}{(\BunboA)(\BunboB)}\\ \end{align*} %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)
【解答】
\newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize\bf\bm #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \dfrac{2}{x^2-1} - \dfrac{1}{x^2+x}\\ & \colMM{red}{分数式➡因数分解!}\\ &= \dfrac{2}{(x+1)\colBX{mistyrose}{$(x-1)$}} - \dfrac{1}{\colBX{palegreen}{$x$}(x+1)}\\ & \colMM{green}{ 相手の分母からもらう \nwarrow} \colMM{orange}{\nearrow 相手の分母からもらう}\\ &= \dfrac{2\colBX{palegreen}{$x$}}{(x+1)\colBX{mistyrose}{$(x-1)$}\colBX{palegreen}{$x$}} - \dfrac{1\colBX{mistyrose}{$(x-1)$}}{\colBX{palegreen}{$x$}(x+1)\colBX{mistyrose}{$(x-1)$}}\\ & \colMM{red}{分母が同じ➡分子だけ計算する}\\ &= \dfrac{2x-(x-1)}{x(x+1)(x-1)}\\ \\ &= \dfrac{\colBX{violet}{$x+1$}}{x\colBX{violet}{$(x+1)$}(x-1)}\\ \\ &= \dfrac{1}{x(x-1)} \end{align*} %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)
【解答】
\def\BunsiA{2} \def\BunboA{x+1} \def\BunsiB{3} \def\BunboB{x-2} \def\Tenkai{2x-4+3x+3} \def\Kotae{5x-1} \newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize\bf\bm #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} &\dfrac{\BunsiA}{\colBX{mistyrose}{$\BunboA$}} + \dfrac{\BunsiB}{\colBX{palegreen}{$\BunboB$}}\\ & \colMM{green}{相手の分母からもらう \swarrow} \colMM{orange}{\searrow 相手の分母からもらう}\\ &= \dfrac{\BunsiA \colBX{palegreen}{$(\BunboB$)}}{\colBX{mistyrose}{$(\BunboA)$}\colBX{palegreen}{$(\BunboB)$}} + \dfrac{\BunsiB\colBX{mistyrose}{$(\BunboA)$}}{\colBX{palegreen}{$(\BunboB)$}\colBX{mistyrose}{$(\BunboA)$}}\\ & \colMM{red}{分母が同じ➡分子だけ計算する}\\ &= \dfrac{\BunsiA(\BunboB)+\BunsiB(\BunboA)}{(\BunboA)(\BunboB)}\\ \\ &= \dfrac{\Tenkai}{(\BunboA)(\BunboB)}\\ \\ &= \dfrac{\Kotae}{(\BunboA)(\BunboB)}\\ \end{align*} %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)
【解答】
\newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize\bf\bm #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \dfrac{x}{x-1} - \dfrac{1}{x^2-x}\\ & \colMM{red}{分数式➡因数分解!}\\ &= \dfrac{x}{x-1} - \dfrac{1}{\colBX{palegreen}{$x$}(x-1)}\\ & \colMM{green}{ \swarrow 相手の分母からもらう}\\% \colMM{orange}{\nearrow 相手の分母からもらう}\\ &= \dfrac{x \times\colBX{palegreen}{$x$}}{(x-1)\colBX{palegreen}{$x$}} - \dfrac{1}{\colBX{palegreen}{$x$}(x-1)}\\ & \colMM{red}{分母が同じ➡分子だけ計算する}\\ &= \dfrac{x^2-1}{x(x-1)}\\ & \colMM{red}{分数式➡因数分解!}\\ &= \dfrac{(x+1)\colBX{violet}{$(x-1)$}}{x\colBX{violet}{$(x-1)$}}\\ & \colMM{red}{\Darr 約分}\\ &= \dfrac{x+1}{x} \end{align*} %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)
【解答】
\newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize\bf\bm #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \dfrac{x}{x+1} + \dfrac{3x-1}{x^2-2x-3}\\ & \colMM{red}{分数式➡因数分解!}\\ &= \dfrac{x}{x+1} + \dfrac{3x-1}{(x+1)\colBX{palegreen}{$(x-3)$}}\\ & \colMM{green}{ \swarrow 相手の分母からもらう}\\% \colMM{orange}{\nearrow 相手の分母からもらう}\\ &= \dfrac{x \times\colBX{palegreen}{$(x-3)$}}{(x+1)\colBX{palegreen}{$(x-3)$}} + \dfrac{3x-1}{(x+1)\colBX{palegreen}{$(x-3)$}}\\ & \colMM{red}{分母が同じ➡分子だけ計算する}\\ &= \dfrac{x(x-3)+(3x-1)}{(x+1)(x-3)}\\ \\ &= \dfrac{x^2-3x+3x-1}{(x+1)(x-3)}\\ \\ &= \dfrac{x^2-1}{(x+1)(x-3)}\\ & \colMM{red}{分数式➡因数分解!}\\ &= \dfrac{\colBX{violet}{$(x+1)$}(x-1)}{\colBX{violet}{$(x+1)$}(x-3)}\\ & \colMM{red}{\Darr 約分}\\ &= \dfrac{x-1}{x-3} \end{align*} %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)
【解答】
\newcommand\colNS[2]{\color{#1}#2\color{black}} \newcommand\colUL[2]{\textcolor{#1}{\underline{\color{black}#2}}} \newcommand\colMM[2]{\color{#1}\scriptsize\bf\bm #2\color{black}} \newcommand\colBX[2]{\colorbox{#1}{#2}} \newcommand\colFR[2]{\textcolor{#1}{\fbox{\color{black}#2}}} \begin{align*} & \dfrac{3x+5}{x^2-1} - \dfrac{1}{x^2+x}\\ & \colMM{red}{分数式➡因数分解!}\\ &= \dfrac{3x+5}{(x+1)\colBX{mistyrose}{$(x-1)$}} - \dfrac{1}{\colBX{palegreen}{$x$}(x+1)}\\ & \colMM{orange}{ 相手の分母からもらう\searrow} \colMM{green}{\swarrow 相手の分母からもらう}\\% \colMM{orange}{\nearrow 相手の分母からもらう}\\ &= \dfrac{(3x+5) \times \colBX{palegreen}{$x$}}{(x+1)\colBX{mistyrose}{$(x-1)$}\colBX{palegreen}{$x$}} - \dfrac{1\colBX{mistyrose}{$(x-1)$}}{\colBX{palegreen}{$x$}(x+1)\colBX{mistyrose}{$(x-1)$}}\\ & \colMM{red}{分母が同じ➡分子だけ計算する}\\ &= \dfrac{x(3x+5)-(x-1)}{x(x+1)(x-1)}\\ \\ &= \dfrac{3x^2+5x-x+1}{x(x+1)(x-1)}\\ \\ &= \dfrac{3x^2+4x+1}{x(x+1)(x-1)}\\ & \colMM{red}{分数式➡因数分解!}\\ &= \dfrac{(3x+1)\colBX{violet}{$(x+1)$}}{x\colBX{violet}{$(x+1)$}(x-1)}\\ & \colMM{red}{\Darr 約分}\\ &= \dfrac{3x+1}{x(x-1)} \end{align*} %1 orange,bisque %2 green,palegreen %3 magenta, violet %4 deepskyblue, lightcyan