数学 - Python - 代数学 - 1次方程式, 2次方程式 - 1次方程式 - 解、移項の法則

学習環境

代数への出発 (新装版数学入門シリーズ) (松坂和夫(著)、岩波書店)の第4章(1次方程式, 2次方程式 )、1(1次方程式)の問1の解答を求めてみる。

1. $x = - 9$
2. $x = - \frac{2}{5}$
3. $\begin{array}{l} 3 x = - 9 \\ x = - 3 \end{array}$
4. $\begin{array}{l} \frac{1}{2} x = 7 \\ x = 14 \end{array}$
5. $\begin{array}{l} 6 x - 3 = 22 + 11 x \\ 5 x = - 25 \\ x = - 5 \end{array}$
6. $\begin{array}{l} \frac{8}{20} x - \frac{5}{20} x = \frac{3}{2} + \frac{6}{2} \\ \frac{3}{20} x = \frac{9}{2} \\ x = \frac{9}{2} \cdot \frac{20}{3} = 30 \end{array}$
7. $x = \frac{1}{\sqrt{2} + 1}$
8. $\begin{array}{l} (\sqrt{2} + \sqrt{3} - 2 \sqrt{2}) x = \sqrt{6} - 2 \\ (\sqrt{3} - \sqrt{2}) x = \sqrt{6} - 2 \\ x = \frac{\sqrt{6} - 2}{\sqrt{3} - \sqrt{2}} \\ = (\sqrt{6} - 2) (\sqrt{3} + \sqrt{2}) \\ = 3 \sqrt{2} + 2 \sqrt{3} - 2 \sqrt{3} - 2 \sqrt{2} \\ = \sqrt{2} \end{array}$
9. $\begin{array}{l} x^{2} - 3 x + 2 = x^{2} - 7 x + 12 \\ 4 x = 10 \\ x = \frac{5}{2} \end{array}$
10. $\begin{array}{l} 18 x^{2} + 7 x - 8 = 18 x^{2} - 3 x - 28 \\ 10 x = - 20 \\ x = - 2 \end{array}$

#!/usr/bin/env python3 from unittest import TestCase, main from sympy import symbols, solve, Rational, sqrt print('1.') class MyTest(TestCase): def test(self): x = symbols('x') eqs = [8 - x - 17, -25 * x - 10, 2 * x - 9 - 5 * x, x / 2 - 3 - 4, 3 * (2 * x - 1) - 11 * (2 + x), 2 * x / 5 - Rational(3, 2) - x / 4 - 3, (sqrt(2) + 1) * x - 1, sqrt(2) * x - sqrt(3) * (sqrt(2) - x) - 2 * (sqrt(2) * x - 1), (x - 1) * (x - 2) - (x - 3) * (x - 4), (2 * x - 1) * (9 * x + 8) - (3 * x - 4) * (6 * x + 7)] xs = [-9, -Rational(2, 5), -3, 14, -5, 30, 1 / (sqrt(2) + 1), sqrt(2), Rational(5, 2), -2] for eq, x0 in zip(eqs, xs): self.assertEqual((solve(eq, x)[0] - x0).simplify(), 0) if __name__ == '__main__': main()

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2020年1月28日火曜日

数学 - Python - 代数学 - 1次方程式, 2次方程式 - 1次方程式 - 解、移項の法則

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