与えられた6つの数式をそれぞれ簡略化する問題です。 (1) $\frac{a^2 + 6ab - 7b^2}{a^2 + 2ab + b^2} \cdot \frac{a+b}{a-b}$ (2) $\frac{p^2 + 5p + 6}{p^2 - 9} \div \frac{p^2 + 4p + 4}{p^3 - 3p^2}$ (3) $\frac{x-3}{x^2-2x} - \frac{x-4}{x^2-4}$ (4) $\frac{a}{ab-b^2} + \frac{b}{ab-a^2}$ (5) $\frac{x}{2x-1} + \frac{x-1}{2x+1} - \frac{4x^2-1}{2x}$ (6) $\left( \frac{x}{x^2-9} - \frac{1}{x-3} \right) \div \frac{x}{x+3}$

1. 問題の内容

与えられた6つの数式をそれぞれ簡略化する問題です。

(1) $\frac{a^2 + 6ab - 7b^2}{a^2 + 2ab + b^2} \cdot \frac{a+b}{a-b}$

(2) $\frac{p^2 + 5p + 6}{p^2 - 9} \div \frac{p^2 + 4p + 4}{p^3 - 3p^2}$

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

(4) $\frac{a}{ab-b^2} + \frac{b}{ab-a^2}$

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

(6) $\left( \frac{x}{x^2-9} - \frac{1}{x-3} \right) \div \frac{x}{x+3}$

2. 解き方の手順

(1)

$a^2 + 6ab - 7b^2 = (a+7b)(a-b)$

$a^2 + 2ab + b^2 = (a+b)^2$

$\frac{a^2 + 6ab - 7b^2}{a^2 + 2ab + b^2} \cdot \frac{a+b}{a-b} = \frac{(a+7b)(a-b)}{(a+b)^2} \cdot \frac{a+b}{a-b} = \frac{a+7b}{a+b}$

(2)

$p^2 + 5p + 6 = (p+2)(p+3)$

$p^2 - 9 = (p+3)(p-3)$

$p^2 + 4p + 4 = (p+2)^2$

$p^3 - 3p^2 = p^2(p-3)$

$\frac{p^2 + 5p + 6}{p^2 - 9} \div \frac{p^2 + 4p + 4}{p^3 - 3p^2} = \frac{(p+2)(p+3)}{(p+3)(p-3)} \div \frac{(p+2)^2}{p^2(p-3)} = \frac{(p+2)(p+3)}{(p+3)(p-3)} \cdot \frac{p^2(p-3)}{(p+2)^2} = \frac{p^2}{p+2}$

(3)

$\frac{x-3}{x^2-2x} - \frac{x-4}{x^2-4} = \frac{x-3}{x(x-2)} - \frac{x-4}{(x+2)(x-2)} = \frac{(x-3)(x+2) - (x-4)x}{x(x-2)(x+2)} = \frac{x^2 - x - 6 - x^2 + 4x}{x(x-2)(x+2)} = \frac{3x-6}{x(x-2)(x+2)} = \frac{3(x-2)}{x(x-2)(x+2)} = \frac{3}{x(x+2)}$

(4)

$\frac{a}{ab-b^2} + \frac{b}{ab-a^2} = \frac{a}{b(a-b)} + \frac{b}{a(b-a)} = \frac{a}{b(a-b)} - \frac{b}{a(a-b)} = \frac{a^2 - b^2}{ab(a-b)} = \frac{(a+b)(a-b)}{ab(a-b)} = \frac{a+b}{ab}$

(5)

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

(6)

$\left( \frac{x}{x^2-9} - \frac{1}{x-3} \right) \div \frac{x}{x+3} = \left( \frac{x}{(x-3)(x+3)} - \frac{1}{x-3} \right) \div \frac{x}{x+3} = \left( \frac{x - (x+3)}{(x-3)(x+3)} \right) \div \frac{x}{x+3} = \frac{-3}{(x-3)(x+3)} \div \frac{x}{x+3} = \frac{-3}{(x-3)(x+3)} \cdot \frac{x+3}{x} = \frac{-3}{x(x-3)}$

3. 最終的な答え

(1) $\frac{a+7b}{a+b}$

(2) $\frac{p^2}{p+2}$

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

(4) $\frac{a+b}{ab}$

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

(6) $\frac{-3}{x(x-3)}$