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(乗算, 除算)

学習環境

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20.を解いてみる。

問20.

$\frac{(x - 7) (x + 7) (x + 2)}{x (x + 2) (x - 7)} = \frac{x + 7}{x}$
$\frac{(x + y) {(x - y)}^{2}}{{(x - y)}^{2} x (x + y)} = \frac{1}{x}$
$\frac{(x + 1) (x + 2) (x + 4) (x - 3)}{(x - 3) (x - 2) (x + 1) (x + 3)} = \frac{(x + 2) (x + 4)}{(x - 2) (x + 3)}$
$\frac{5 (x - 1) (x + 2) (x^{2} - 2 x + 4)}{(x - 6) (x + 2) 4 (x + 1) (x - 1)} = \frac{5 (x^{2} - 2 x + 4)}{4 (x - 6) (x + 1)}$
$\frac{(a - 2) (x - 3) 2 (a + 1) (3 a + 2)}{(a - 1) (3 a + 2) (a - 3) (a + 2)} = \frac{2 (a - 2) (a + 1)}{(a - 1) (a + 2)}$
$\frac{(1 + a) (1 - a) (1 - b) (1 + b)}{(1 + b) a (1 + a) (1 - a)} = \frac{1 - b}{a}$
$\frac{(2 x - 5) (3 x + 4) (x - 2) (x + 1) (x + 2) x}{(x + 2) (x - 2) 3 x (2 x - 5) (x + 1) (3 x + 4)} = \frac{1}{3}$
$\begin{array}{l} \frac{(x - 1) (x - 3) - 4 (x - 3) + 12 (x - 1)}{(x - 1) (x - 3)} \cdot \frac{(x + 1) (x + 3) + 4 (x + 3) - 12 (x + 1)}{(x + 1) (x + 3)} \\ = \frac{(x^{2} + 4 x + 3) (x^{2} - 4 x + 3)}{(x - 1) (x - 3) (x + 1) (x + 3)} \\ = \frac{(x + 1) (x + 3) (x - 1) (x - 3)}{(x - 1) (x - 3) (x + 1) (x + 3)} \\ = 1 \end{array}$

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