数学 – Python - 数学はここから始まる - 数 – 分数式の演算と分数式 – 分数式の演算(加法・減法(和と差)、既約、通分、因数分解)

数学読本〈1〉

学習環境

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• Windows 10 Pro (OS)
• Nebo(Windows アプリ)
• iPad Pro + Apple Pencil
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• 参考書籍
  • Pythonからはじめる数学入門
  • JavaScriptリファレンス 第6版
  • インタラクティブ・データビジュアライゼーション

数学読本〈1〉数・式の計算/方程式/不等式 (松坂 和夫(著)、岩波書店)の第2章(文字と記号の活躍 - 式の計算)、2.3(分数式の演算と分数式)、分数式の演算の問19.を取り組んでみる。

19. 
    
    1. 
       
       $\frac{ac-bc+ab-ac+bc-ab}{abc}=0$
    2. 
       
       $\frac{x-1}{x^2-1}=\frac{1}{x+1}$
    3. 
       
       $\frac{x+2-4}{x-2}=1$
    4. 
       
       $\begin{array}{}\frac{x-1-2x}{\left(x+1\right)\left(x-1\right)}\\ =\frac{-x-1}{\left(x+1\right)\left(x-1\right)}\\ =-\frac{1}{x-1}\end{array}$
    5. 
       
       $\begin{array}{}\frac{x^2+3x+2-2x}{x+1}-\frac{3x^2+4}{x\left(x+1\right)}\\ =\frac{x^2+x+2}{x+1}-\frac{3x^2+4}{x\left(x+1\right)}\\ =\frac{x^3+x^2+2x-3x^2-4}{x\left(x+1\right)}\\ =\frac{x^3-2x^2+2x-4}{x\left(x+1\right)}\\ =\frac{\left(x-2\right)\left(x^2+2\right)}{x\left(x+1\right)}\end{array}$
    6. 
       
       $\begin{array}{}\frac{x^2-x-2-x^2+x-1+x^2+x+3}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\frac{x^2+x}{\left(x+1\right)\left(x^2-x+1\right)}\\ =\frac{x}{x^2-x+1}\end{array}$
    7. 
       
       $\begin{array}{}\frac{1}{\left(x-1\right)\left(x-2\right)}+\frac{2}{\left(2x+1\right)\left(x-1\right)}-\frac{3}{\left(2x+1\right)\left(x-2\right)}\\ =\frac{2x+1+2x-4-3x+3}{\left(x-1\right)\left(x-2\right)\left(2x+1\right)}\\ =\frac{x}{\left(x-1\right)\left(x-2\right)\left(2x+1\right)}\end{array}$
    8. 
       
       $\begin{array}{}\frac{1}{x\left(x+1\right)}+\frac{-x-3+x+2}{\left(x+2\right)\left(x+3\right)}\\ =\frac{1}{x\left(x+1\right)}-\frac{1}{\left(x+2\right)\left(x+3\right)}\\ =\frac{x^2+5x+6-x^2-x}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\\ =\frac{2\left(2x+3\right)}{x\left(x+1\right)\left(x+2\right)\left(x+3\right)}\end{array}$
    9. 
       
       $\begin{array}{}\frac{2a}{a^2-1}+\frac{2a}{a^2+1}+\frac{4a^3}{a^4+1}\\ =\frac{4a^3}{a^4-1}+\frac{4a^3}{a^4+1}\\ =\frac{8a^7}{a^8-1}\\ =\frac{8a^7}{\left(a+1\right)\left(a-1\right)\left(a^2+1\right)\left(a^4+1\right)}\end{array}$
    10. 
        
        $\begin{array}{}\frac{x+2+x}{x\left(x+1\right)\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\\ =\frac{2}{x\left(x+2\right)}+\frac{1}{\left(x+2\right)\left(x+3\right)}\\ =\frac{2x+6+x}{x\left(x+2\right)\left(x+3\right)}\\ =\frac{3}{x\left(x+3\right)}\end{array}$
    11. 
        
        $\frac{-a\left(b-c\right)-b\left(c-a\right)-c\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0$
    12. 
        
        $\begin{array}{}\frac{-\left(b-c\right)\left(b+1\right)-\left(c-a\right)\left(a+1\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+1\right)\left(b+1\right)}\\ =\frac{-b^2-b+bc+c-ca-c+a^2+a}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+1\right)\left(b+1\right)}\\ =\frac{-b^2-b+bc+a^2-ca+a}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+1\right)\left(b+1\right)}\\ =\frac{\left(a+b\right)\left(a-b\right)+\left(1-c\right)\left(a-b\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)\left(a+1\right)\left(b+1\right)}\\ =\frac{a+b-c+1}{\left(b-c\right)\left(c-a\right)\left(a+1\right)\left(b+1\right)}\end{array}$

        残りの計算。

        $\frac{\left(a+b-c+1\right)\left(c+1\right)-\left(a+1\right)\left(b+1\right)}{\left(b-c\right)\left(c-a\right)\left(a+1\right)\left(b+1\right)\left(c+1\right)}$

        分子の計算。

        $\begin{array}{}-c^2+\left(a+b\right)c+a+b+1-ab-a-b-1\\ =-c^2+\left(a+b\right)c-ab\\ =-\left(c^2-\left(a+b\right)c+ab\right)\\ =-\left(c-a\right)\left(c-b\right)\\ =\left(b-c\right)\left(c-a\right)\end{array}$

        よって求める計算結果は

        $\frac{1}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}$

コード(Emacs)

Python 3

#!/usr/bin/env python3
from sympy import pprint, symbols, gcd, lcm

print('19.')

a, b, c, x = symbols('a, b, c, x')

ts = [(a - b) / (a * b) + (b - c) / (b * c) + (c - a) / (c * a),
      x / (x ** 2 - 1) - 1 / (x ** 2 - 1),
      (x + 2) / (x - 2) + 4 / (2 - x),
      1 / (x + 1) - 2 * x / (x ** 2 - 1),
      x + 2 - 2 * x / (x + 1) - (3 * x ** 2 + 4) / (x * (x + 1)),
      (x - 2) / (x ** 2 - x + 1) - 1 /
      (x + 1) + (x ** 2 + x + 3) / (x ** 3 + 1),
      1 / (x ** 2 - 3 * x + 2) + 2 / (2 * x ** 2 - x - 1) -
      3 / (2 * x ** 2 - 3 * x - 2),
      1 / x - 1 / (x + 1) - 1 / (x + 2) + 1 / (x + 3),
      1 / (a - 1) + 1 / (a + 1) + 2 * a /
      (a ** 2 + 1) + 4 * a ** 3 / (a ** 4 + 1),
      1 / (x * (x + 1)) + 1 / ((x + 1) * (x + 2)) + 1 / ((x + 2) * (x + 3)),
      a / ((a - b) * (a - c)) + b /
      ((b - c) * (b - a)) + c / ((c - a) * (c - b)),
      1 / ((a - b) * (a - c) * (a + 1)) + 1 / ((b - c) * (b - a) * (b + 1)) + 1 / ((c - a) * (c - b) * (c + 1))]

for i, t in enumerate(ts, 1):
    print(f'({i})')
    pprint(t.factor())
    print()

入出力結果(Terminal, Jupyter(IPython))

$ ./sample19.py
19.
(1)
0

(2)
  1  
─────
x + 1

(3)
1

(4)
 -1  
─────
x - 1

(5)
        ⎛ 2    ⎞
(x - 2)⋅⎝x  + 2⎠
────────────────
   x⋅(x + 1)    

(6)
    x     
──────────
 2        
x  - x + 1

(7)
            x            
─────────────────────────
(x - 2)⋅(x - 1)⋅(2⋅x + 1)

(8)
       2⋅(2⋅x + 3)       
─────────────────────────
x⋅(x + 1)⋅(x + 2)⋅(x + 3)

(9)
                  7              
               8⋅a               
─────────────────────────────────
                ⎛ 2    ⎞ ⎛ 4    ⎞
(a - 1)⋅(a + 1)⋅⎝a  + 1⎠⋅⎝a  + 1⎠

(10)
    3    
─────────
x⋅(x + 3)

(11)
0

(12)
           1           
───────────────────────
(a + 1)⋅(b + 1)⋅(c + 1)

$