数学 - 代数学 - 因数分解と分数式 - 整式の加法、乗法、除法、通分、最小公倍数、因数分解、既約

代数への出発
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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3章(因数分解と分数式)、練習問題の問3の解答を求めてみる。

1. $\begin{array}{l} \frac{1}{x} - \frac{1}{x + 1} - \frac{1}{x + 2} + \frac{1}{x + 3} \\ = \frac{1}{x (x + 1)} - \frac{1}{x + 2} + \frac{1}{x + 3} \\ = \frac{x + 2 - x^{2} - x}{x (x + 1) (x + 2)} + \frac{1}{x + 3} \\ = \frac{2 - x^{2}}{x (x + 1) (x + 2)} + \frac{1}{x + 3} \\ = \frac{- x^{3} - 3 x^{2} + 2 x + 6 + x^{3} + 3 x^{2} + 2 x}{x (x + 1) (x + 2) (x + 3)} \\ = \frac{4 x + 6}{x (x + 1) (x + 2) (x + 3)} \end{array}$
2. $\begin{array}{l} \frac{1}{x (x + 1)} + \frac{1}{(x + 1) (x + 2)} + \frac{1}{(x + 2) (x + 3)} \\ = \frac{(x + 2) (x + 3) + x (x + 3) + x (x + 1)}{x (x + 1) (x + 2) (x + 3)} \\ = \frac{x^{2} + 5 x + 6 + x^{2} + 3 x + x^{2} + x}{x (x + 1) (x + 2) (x + 3)} \\ = \frac{3 x^{2} + 2 x^{2} + 6 x + 6}{x (x + 1) (x + 2) (x + 3)} \end{array}$
3. $\begin{array}{l} \frac{a + 2 b}{a (a - b)} + \frac{2 a + b}{(a - b) (a - 2 b)} + \frac{a - 3 b}{a (a - 2 b)} \\ = \frac{(a + 2 b) (a - 2 b) + (2 a + b) a + (a - 3 b) (a - b)}{a (a - b) (a - 2 b)} \\ = \frac{a^{2} - 4 b^{2} + 2 a^{2} + a b + a^{2} - 4 a b + 3 b^{2}}{a (a - b) (a - 2 b)} \\ = \frac{4 a^{2} - 3 a b - b^{2}}{a (a - b) (a - 2 b)} \\ = \frac{(a - b) (4 a + b)}{a (a - b) (a - 2 b)} \\ = \frac{4 a + b}{a (a - 2 b)} \end{array}$
4. $\begin{array}{l} \frac{1 - (a + 4) + a + 3}{(a + 2) (a + 3) (a + 4)} \\ = 0 \end{array}$
5. $\begin{array}{l} \frac{(x + y - z) (x - y + z)}{(x + y + z) (x + y - z)} + \frac{(y + z - x) (y - z + x)}{(y + z + x) (y + z - x)} \\ + \frac{(z + x - y) (z - x + y)}{(z + x + y) (z + x - y)} \\ = \frac{x - y + z}{x + y + z} + \frac{y - z + x}{y + z + x} + \frac{z - x + y}{z + x + y} \\ = \frac{x + y + z}{x + y + z} \\ = 1 \end{array}$
6. $\begin{array}{l} 1 + \frac{1}{x + 1} - 1 - \frac{1}{x + 2} + 1 + \frac{1}{x + 3} - 1 - \frac{1}{x + 4} \\ = \frac{1}{x + 1} - \frac{1}{x + 2} + \frac{1}{x + 3} - \frac{1}{x + 4} \\ = \frac{1}{(x + 1) (x + 2)} + \frac{1}{x + 3} - \frac{1}{x + 4} \\ = \frac{x + 3 + x^{2} + 3 x + 2}{(x + 1) (x + 2) (x + 3)} - \frac{1}{x + 4} \\ = \frac{x^{2} + 4 x + 5}{(x + 1) (x + 2) (x + 3)} - \frac{1}{x + 4} \\ = \frac{x^{3} + 8 x^{2} + 21 x + 20 - (x^{2} + 3 x + 2) (x + 3)}{(x + 1) (x + 2) (x + 3) (x + 4)} \\ = \frac{x^{3} + 8 x^{2} + 21 x + 20 - x^{3} - 6 x^{2} - 12 x - 6}{(x + 1) (x + 2) (x + 3) (x + 4)} \\ = \frac{2 x^{2} + 9 x + 14}{(x + 1) (x + 2) (x + 3) (x + 4)} \end{array}$
7. $\begin{array}{l} \frac{(2 a + 1) (2 a - 1)}{(a - 2) (a - 4)} \cdot \frac{(a - 4) (a + 2)}{(2 a + 1) (a + 2)} \\ = \frac{2 a - 1}{a - 2} \end{array}$
8. $\begin{array}{l} \frac{(2 x - 1) (x - 2)}{(x + 6) (x - 5)} \cdot \frac{(x + 2) (x + 6)}{(2 x - 1) (x + 5)} \\ = \frac{(x - 2) (x + 2)}{(x - 5) (x + 5)} \\ = \frac{x^{2} - 4}{x^{2} - 25} \end{array}$
9. $\begin{array}{l} \frac{(x + 4) (x - 4)}{{(x^{2} + 4)}^{2} - 4 x^{2}} \cdot \frac{(x - 2) (x^{2} + 2 x + 4)}{x^{2} + 4} \cdot \frac{x^{2} - 2 x + 4}{{(x - 2)}^{2}} \cdot \frac{1}{x + 2} \\ = \frac{(x + 4) (x - 4) (x^{2} + 2 x + 4) (x^{2} - 2 x + 4)}{(x^{2} + 2 x + 4) (x^{2} - 2 x + 4) (x^{2} + 4) (x - 2) (x + 2)} \\ = \frac{x^{2} - 16}{x^{4} - 16} \end{array}$
10. $\begin{array}{l} \frac{(x + y) (x - y) (x - y + z) (x - y - z) x (x + y - z)}{(x + y - z) (x - y + z) x (x - y) {(x + y)}^{2}} \\ = \frac{x - y - z}{x + y} \end{array}$
11. $\begin{array}{l} \frac{x^{2} - 4 x + 3 - 4 x + 12 + 12 x - 12}{(x - 1) (x - 3)} \cdot \frac{x^{2} + 4 x + 3 + 4 x + 12 - 12 x - 12}{(x + 1) (x + 3)} \\ = \frac{x^{2} + 4 x + 3}{(x - 1) (x - 3)} \cdot \frac{x^{2} - 4 x + 3}{(x + 1) (x + 3)} \\ = 1 \end{array}$
12. $\begin{array}{l} \frac{{(a - b)}^{3} + {(b - c)}^{3} + {(c - a)}^{3}}{(a - b) (b - c) (c - a)} \\ = \frac{((a - b) + (b - c) + (c - a)) ({(a - b)}^{2} + {(b - c)}^{2} + {(c - a)}^{2}) + 3 (a - b) (b - c) (c - a)}{(a - b) (b - c) (c - a)} \\ = 3 \end{array}$
13. $\begin{array}{l} \frac{- (b - c) (b + 1) (c + 1) - (c - a) (c + 1) (a + 1) - (a - b) (a + 1) (b + 1)}{(a - b) (b - c) (c - a) (a + 1) (b + 1) (c + 1)} \\ = \frac{(c + 1 - b - 1) a^{2} + (- c (c + 1) + c + 1 - (b + 1) + b (b + 1)) a}{(a - b) (b - c) (c - a) (a + 1) (b + 1) (c + 1)} \\ + \frac{- (b - c) (b + 1) (c + 1) - c (c + 1) + b (b + 1)}{(a - b) (b - c) (c - a) (a + 1) (b + 1) (c + 1)} \\ = \frac{(c - b) a^{2} + (b^{2} - c^{2}) a + (1 - c - 1) b^{2} + (1 - c - 1 + c (c + 1)) b + c (c + 1) - c (c + 1)}{(a - b) (b - c) (c - a) (a + 1) (b + 1) (c + 1)} \\ = \frac{- (b - c) a^{2} + (b - c) (b + c) a - c b^{2} + c^{2} b}{(a - b) (b - c) (c - a) (a + 1) (b + 1) (c + 1)} \\ = \frac{- (b - c) a^{2} + (b - c) (b + c) a - b c (b - c)}{(α - b) (b - c) (c - a) (a + 1) (b + 1) (c + 1)} \\ = \frac{- a^{2} + (b + c) a - b c}{(a - b) (c - a) (a + 1) (b + 1) (c + 1)} \\ = \frac{- (a - b) (a - c)}{(a - b) (c - a) (a + 1) (b + 1) (c + 1)} \\ = \frac{1}{(a + 1) (b + 1) (c + 1)} \end{array}$
14. $\begin{array}{l} 1 - \frac{1}{1 - \frac{1}{1 - \frac{1}{\frac{x - 1}{x}}}} \\ = 1 - \frac{1}{1 - \frac{1}{1 - \frac{x}{x - 1}}} \\ = 1 - \frac{1}{1 - \frac{1}{\frac{- 1}{x - 1}}} \\ = 1 - \frac{1}{1 + x - 1} \\ = 1 - \frac{1}{x} \\ = \frac{x - 1}{x} \end{array}$
15. $\begin{array}{l} \frac{\frac{x + 6}{x + 2}}{x^{2} - 10 - \frac{x^{3} + 8}{\frac{x^{2} + 4 x + 4}{x + 4}}} \\ = \frac{x + 6}{x + 2} \cdot \frac{1}{x^{2} - 10 - \frac{(x + 2) (x^{2} - 2 x + 4) (x + 4)}{{(x + 2)}^{2}}} \\ = \frac{x + 6}{x + 2} \cdot \frac{1}{x^{2} - 10 - \frac{(x^{2} - 2 x + 4) (x + 4)}{x + 2}} \\ = \frac{x + 6}{(x^{2} - 10) (x + 2) - (x^{3} + 2 x^{2} - 4 x + 16)} \\ = \frac{x + 6}{- 6 x - 36} \\ = \frac{x + 6}{- 6 (x + 6)} \\ = - \frac{1}{6} \end{array}$

Kamimura's blog

2020年1月21日火曜日

数学 - 代数学 - 因数分解と分数式 - 整式の加法、乗法、除法、通分、最小公倍数、因数分解、既約

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