$a = \frac{1}{2}$ のとき、以下の式の値を求める問題です。 (1) $a^2$ (2) $-a^2$ (3) $3a^2 - 2a^2$ (4) $4a^2 - 5a^2$ (5) $(-a)^2$ (6) $-(-a)^2$ (7) $(2a)^2$ (8) $(-2a)^2$ (9) $-(2a)^2$ (10) $-(-2a)^2$

1. 問題の内容

$a = \frac{1}{2}$ のとき、以下の式の値を求める問題です。

(1) $a^2$

(2) $-a^2$

(3) $3a^2 - 2a^2$

(4) $4a^2 - 5a^2$

(5) $(-a)^2$

(6) $-(-a)^2$

(7) $(2a)^2$

(8) $(-2a)^2$

(9) $-(2a)^2$

(10) $-(-2a)^2$

2. 解き方の手順

$a = \frac{1}{2}$ を各々の式に代入して計算します。

(1) $a^2 = (\frac{1}{2})^2 = \frac{1}{4}$

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

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

(4) $4a^2 - 5a^2 = -a^2 = -(\frac{1}{2})^2 = -\frac{1}{4}$

(5) $(-a)^2 = (-\frac{1}{2})^2 = \frac{1}{4}$

(6) $-(-a)^2 = -(-\frac{1}{2})^2 = -\frac{1}{4}$

(7) $(2a)^2 = (2 \cdot \frac{1}{2})^2 = (1)^2 = 1$

(8) $(-2a)^2 = (-2 \cdot \frac{1}{2})^2 = (-1)^2 = 1$

(9) $-(2a)^2 = -(2 \cdot \frac{1}{2})^2 = -(1)^2 = -1$

(10) $-(-2a)^2 = -(-2 \cdot \frac{1}{2})^2 = -(-1)^2 = -1$

3. 最終的な答え

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

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

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

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

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

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

(7) $1$

(8) $1$

(9) $-1$

(10) $-1$