21102中２・式の計算・計算問題・分数をふくむ式2

$\begin{array}{r}(1){\displaystyle \frac{x}{3}}+{\displaystyle \frac{x+y}{4}}-{\displaystyle \frac{2x-3y}{6}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{4x}{12}}+{\displaystyle \frac{3(x+y)}{12}}-{\displaystyle \frac{2(2x-3y)}{12}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{4x+3(x+y)-2(2x-3y)}{12}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{4x+3x+3y-4x+6y}{12}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{4x+3x-4x+3y+6y}{12}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{{\cancel{3}}^{1}x+{\cancel{9}}^{3}y}{{\cancel{12}}^4}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{x+3y}{4}}\end{array}$

答${\displaystyle \frac{x+3y}{4}}$

$\begin{array}{r}(2){\displaystyle \frac{3x+6y}{5}}+{\displaystyle \frac{5x-3y}{4}}-{\displaystyle \frac{6x-3y}{10}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{4(3x+6y)}{20}}+{\displaystyle \frac{5(5x-3y)}{20}}-{\displaystyle \frac{2(6x-3y)}{20}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{4(3x+6y)+5(5x-3y)-2(6x-3y)}{20}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{12x+24y+25x-15y-12x+6y}{20}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{12x+25x-12x+24y-15y+6y}{20}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{{\cancel{25}}^{5}x+{\cancel{15}}^{3}y}{{\cancel{20}}^4}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{5x+3y}{4}}\end{array}$

答${\displaystyle \frac{5x+3y}{4}}$

$\begin{array}{r}(3){\displaystyle \frac{4a-3}{5}}-{\displaystyle \frac{5b+3}{2}}-3a+2b\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{2(4a-3)}{10}}-{\displaystyle \frac{5(5b+3)}{10}}-{\displaystyle \frac{3a\times 10}{10}}+{\displaystyle \frac{2b\times 10}{10}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{2(4a+3)-5(5b+3)-30a+20b}{10}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{8a+6-25b-15-30a+20b}{10}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{8a-30a-25b+20b+6-15}{10}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{-22a-5b-9}{10}}\end{array}$

答${\displaystyle \frac{-22a-5b-9}{10}}$

$\begin{array}{r}(4)2(x^2-3x-6)-{\displaystyle \frac{3x^2-4x+5}{2}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{2\times 2(x^2-3x-6)}{2}}-{\displaystyle \frac{3x^2-4x+5}{2}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{4(x^2-3x-6)-(3x^2-4x+5)}{2}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{4x^2-12x-24-3x^2+4x-5}{2}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{4x^2-3x^2-12x+4x-24-5}{2}}\end{array}$

$\begin{array}{rcl}& =& {\displaystyle \frac{x^2-8x-29}{2}}\end{array}$

答${\displaystyle \frac{x^2-8x-29}{2}}$