数学 – 文字と記号の活躍 - 整式の除法と分数式 – 分数式の演算

数学読本〈1〉
数・式の計算/方程式/不等式
(岩波書店)
松坂和夫 (著)

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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(整式の除法と分数式)、部sん数式の問19.を解いてみる。

問19.

$\frac{a c - b c + a b - a c + b c - a b}{a b c} = 0$
$\frac{x - 1}{x^{2} - 1} = \frac{1}{x + 1}$
$\frac{x - 1}{x^{2} - 1} = \frac{1}{x + 1}$
$\frac{x - 1 - 2 x}{x^{2} - 1} = \frac{- 1}{x - 1}$
$\begin{array}{l} \frac{(x + 2) x (x + 1) - 2 x^{2} - 3 x^{2} - 4}{x (x + 1)} \\ = \frac{x^{3} + 3 x^{2} + 2 x - 2 x^{2} - 3 x^{2} - 4}{x (x + 1)} \\ = \frac{x^{3} - 2 x^{2} + 2 x - 4}{x (x + 1)} \\ = \frac{(x^{2} + 2) (x - 2)}{x (x + 1)} \end{array}$
$\begin{array}{l} \frac{(x + 2) x (x + 1) - 2 x^{2} - 3 x^{2} - 4}{x (x + 1)} \\ = \frac{x^{3} + 3 x^{2} + 2 x - 2 x^{2} - 3 x^{2} - 4}{x (x + 1)} \\ = \frac{x^{3} - 2 x^{2} + 2 x - 4}{x (x + 1)} \\ = \frac{(x^{2} + 2) (x - 2)}{x (x + 1)} \end{array}$
$\begin{array}{l} \frac{1}{(x - 2) (x - 1)} + \frac{2}{(2 x + 1) (x - 1)} + \frac{- 3}{(2 x + 1) (x - 2)} \\ = \frac{2 x + 1 + 2 x - 4 - 3 x + 3}{(2 x + 1) (x - 1) (x - 2)} \\ = \frac{x}{(2 x + 1) (x - 1) (x - 2)} \\ \frac{(x + 1) (x + 2) (x + 3) - x (x + 2) (x + 3) - x (x + 1) (x + 3) + x (x + 1) (x + 2)}{x (x + 1) (x + 2) (x + 3)} \end{array}$
$\begin{array}{l} \frac{(x + 1) (x + 2) (x + 3) - x (x + 2) (x + 3) - x (x + 1) (x + 3) + x (x + 1) (x + 2)}{x (x + 1) (x + 2) (x + 3)} \\ = \frac{(x + 2) (x + 3) - x (x + 1)}{x (x + 1) (x + 2) (x + 3)} \\ = \frac{4 x + 6}{x (x + 1) (x + 2) (x + 3)} \\ = \frac{2 (2 x + 3)}{x (x + 1) (x + 2) (x + 3)} \end{array}$
$\begin{array}{l} \frac{2 a}{a^{2} - 1} + \frac{2 a}{a^{2} + 1} + \frac{4 a^{3}}{a^{4} + 1} \\ = \frac{4 a^{3}}{a^{4} - 1} + \frac{4 a^{3}}{a^{4} + 1} \\ = \frac{8 a^{7}}{a^{8} - 1} \end{array}$
$\begin{array}{l} \frac{2 x + 2}{x (x + 1) (x + 2)} + \frac{1}{(x + 2) (x + 3)} \\ = \frac{2}{x (x + 2)} + \frac{1}{(x + 2) (x + 3)} \\ = \frac{2 x + 6 + x}{x (x + 2) (x + 3)} \\ = \frac{3}{x (x + 3)} \end{array}$
$\frac{- a (b - c) - b (c - a) - c (a - b)}{(a - b) (b - c) (c - a)} = 0$
$\begin{array}{l} \frac{- (b + 1) (c + 1) (b - c) - (c + 1) (a + 1) (c - a) - (a + 1) (b + 1) (a - b)}{(a - b) (b - c) (c - a) (a + 1) (b + 1) (c + 1)} \\ = \frac{((c + 1) - (b + 1)) a^{2} + ((c + 1) - (c + 1) c + (b + 1) b - (b + 1)) a + (- (b + 1) (c + 1) (b - c) - (c + 1) c + (b + 1) b)}{(a - b) (b - c) (c - a) (a + 1) (b + 1) (c + 1)} \\ = \frac{(c - b) a^{2} + (1 - c^{2} + b^{2} - 1) a + (- c b^{2} + ((c + 1) c - c - 1 + 1) b + (c + 1) c - (c + 1) c)}{(a - b) (b - c) (c - a) (a + 1) (b + 1) (c + 1)} \\ = \frac{- (b - c) a^{2} + (b^{2} - c^{2}) a + (- c b^{2} + c^{2} b)}{(a - b) (b - c) (c - a) (a + 1) (b + 1) (c + 1)} \\ = \frac{- (b - c) a^{2} + (b - c) (b + c) a - b c (b - c)}{(a - b) (b - c) (c - a) (a + 1) (b + 1) (c + 1)} \\ = \frac{- a^{2} + (b + c) a - b c}{(a - b) (c - a) (a + 1) (b + 1) (c + 1)} \\ = \frac{(a - b) (c - a)}{(a - b) (c - a) (a + 1) (b + 1) (c + 1)} \\ = \frac{1}{(a + 1) (b + 1) (c + 1)} \end{array}$

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2015年3月6日金曜日

数学 – 文字と記号の活躍 - 整式の除法と分数式 – 分数式の演算

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