与えられた6つの数式をそれぞれ簡略化します。 (1) $\frac{16xy^2z}{12x^3yz^4}$ (2) $\frac{x+y}{x-y} - \frac{x^2+y^2}{x^2-y^2}$ (3) $\frac{1}{x} - \frac{y}{x(x+y)} + \frac{x}{y(x+y)}$ (4) $\frac{a^2-5a+6}{a^2-7a+12} \times \frac{a^2-16}{a^2-4} \div \frac{a+4}{a+2}$ (5) $\frac{x - \frac{2}{x+1}}{x+1-\frac{4}{x+1}}$ (6) $\frac{\frac{1}{x-1}+1}{\frac{1}{x+1}-1}$

1. 問題の内容

与えられた6つの数式をそれぞれ簡略化します。

(1) $\frac{16xy^2z}{12x^3yz^4}$

(2) $\frac{x+y}{x-y} - \frac{x^2+y^2}{x^2-y^2}$

(3) $\frac{1}{x} - \frac{y}{x(x+y)} + \frac{x}{y(x+y)}$

(4) $\frac{a^2-5a+6}{a^2-7a+12} \times \frac{a^2-16}{a^2-4} \div \frac{a+4}{a+2}$

(5) $\frac{x - \frac{2}{x+1}}{x+1-\frac{4}{x+1}}$

(6) $\frac{\frac{1}{x-1}+1}{\frac{1}{x+1}-1}$

2. 解き方の手順

(1) 分数式の簡略化

$\frac{16xy^2z}{12x^3yz^4} = \frac{4y}{3x^2z^3}$

(2) 分数式の引き算

$\frac{x+y}{x-y} - \frac{x^2+y^2}{x^2-y^2} = \frac{(x+y)(x+y)}{x^2-y^2} - \frac{x^2+y^2}{x^2-y^2} = \frac{x^2+2xy+y^2 - (x^2+y^2)}{x^2-y^2} = \frac{2xy}{x^2-y^2}$

(3) 分数式の足し引き

$\frac{1}{x} - \frac{y}{x(x+y)} + \frac{x}{y(x+y)} = \frac{y(x+y) - y^2 + x^2}{xy(x+y)} = \frac{xy + y^2 - y^2 + x^2}{xy(x+y)} = \frac{x^2+xy}{xy(x+y)} = \frac{x(x+y)}{xy(x+y)} = \frac{1}{y}$

(4) 分数式の掛け算と割り算

$\frac{a^2-5a+6}{a^2-7a+12} \times \frac{a^2-16}{a^2-4} \div \frac{a+4}{a+2} = \frac{(a-2)(a-3)}{(a-3)(a-4)} \times \frac{(a-4)(a+4)}{(a-2)(a+2)} \times \frac{a+2}{a+4} = 1$

(5) 複合分数式

$\frac{x - \frac{2}{x+1}}{x+1-\frac{4}{x+1}} = \frac{\frac{x(x+1)-2}{x+1}}{\frac{(x+1)^2-4}{x+1}} = \frac{x^2+x-2}{(x+1)^2-4} = \frac{(x+2)(x-1)}{x^2+2x+1-4} = \frac{(x+2)(x-1)}{x^2+2x-3} = \frac{(x+2)(x-1)}{(x+3)(x-1)} = \frac{x+2}{x+3}$

(6) 複合分数式

$\frac{\frac{1}{x-1}+1}{\frac{1}{x+1}-1} = \frac{\frac{1+x-1}{x-1}}{\frac{1-(x+1)}{x+1}} = \frac{\frac{x}{x-1}}{\frac{-x}{x+1}} = \frac{x}{x-1} \times \frac{x+1}{-x} = -\frac{x+1}{x-1}$

3. 最終的な答え

(1) $\frac{4y}{3x^2z^3}$

(2) $\frac{2xy}{x^2-y^2}$

(3) $\frac{1}{y}$

(4) $1$

(5) $\frac{x+2}{x+3}$

(6) $-\frac{x+1}{x-1}$