$x = -\frac{1}{2}$ のとき、以下の各式について、その値を求めよ。 (1) $2x - 5 + 3x$ (2) $\frac{2}{3}x - \frac{1}{4}$ (3) $\frac{2x + 4}{3}$ (4) $\frac{-4x - 5}{7}$ (5) $-6(x - 2)$ (6) $-12(-3x + 1)$ (7) $x^2$ (8) $-x^2$ (9) $-(-x)^2$

1. 問題の内容

$x = -\frac{1}{2}$ のとき、以下の各式について、その値を求めよ。

(1) $2x - 5 + 3x$

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

(3) $\frac{2x + 4}{3}$

(4) $\frac{-4x - 5}{7}$

(5) $-6(x - 2)$

(6) $-12(-3x + 1)$

(7) $x^2$

(8) $-x^2$

(9) $-(-x)^2$

2. 解き方の手順

与えられた $x = -\frac{1}{2}$ を各式に代入して計算する。

(1) $2x - 5 + 3x = 5x - 5$

$5x - 5 = 5(-\frac{1}{2}) - 5 = -\frac{5}{2} - 5 = -\frac{5}{2} - \frac{10}{2} = -\frac{15}{2}$

(2) $\frac{2}{3}x - \frac{1}{4} = \frac{2}{3}(-\frac{1}{2}) - \frac{1}{4} = -\frac{1}{3} - \frac{1}{4} = -\frac{4}{12} - \frac{3}{12} = -\frac{7}{12}$

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

(4) $\frac{-4x - 5}{7} = \frac{-4(-\frac{1}{2}) - 5}{7} = \frac{2 - 5}{7} = \frac{-3}{7} = -\frac{3}{7}$

(5) $-6(x - 2) = -6(-\frac{1}{2} - 2) = -6(-\frac{1}{2} - \frac{4}{2}) = -6(-\frac{5}{2}) = 15$

(6) $-12(-3x + 1) = -12(-3(-\frac{1}{2}) + 1) = -12(\frac{3}{2} + 1) = -12(\frac{3}{2} + \frac{2}{2}) = -12(\frac{5}{2}) = -30$

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

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

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

3. 最終的な答え

(1) $-\frac{15}{2}$

(2) $-\frac{7}{12}$

(3) $1$

(4) $-\frac{3}{7}$

(5) $15$

(6) $-30$

(7) $\frac{1}{4}$

(8) $-\frac{1}{4}$

(9) $-\frac{1}{4}$