$$m, n \ \text{が整数のとき}$$ $$a^0 = 1, \quad a^1 = a, \quad a^m \times a^n = a^{m+n}$$

$$(a + b)^2 = a^2 + 2ab + b^2$$ $$(a - b)^2 = a^2 - 2ab + b^2$$ $$(a + b)(a - b) = a^2 - b^2$$ $$(x + a)(x + b) = x^2 + (a+b)x + ab$$ $$(ax + b)(cx + d) = acx^2 + (ad + bc)x + bd$$ $$(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$$ $$(a - b)^2 = a^3 - 3a^2b + 3ab^2 - b^3$$ $$(a + b)(a^2 - ab + b^2) = a^3 + b^3$$ $$(a - b)(a^2 + ab + b^2) = a^3 - b^3$$

$$a^2 + 2ab + b^2 = (a + b)^2$$ $$a^2 - 2ab + b^2 = (a - b)^2$$ $$a^2 - b^2 = (a + b)(a - b)$$ $$acx^2 + (ad + bc)x + bd = (ax + b)(cx + d)$$ $$$$

$$a^3 + b^3 = (a + b)(a^2 - ab + b^2)$$ $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$

$$a \ge 0 \ \text{のとき} \quad |a| = a$$ $$a \lt 0 \ \text{のとき} \quad |a| = -a$$

$$a, b \ \text{を実数とすると}$$ $$|a| \ge 0, |a| = 0 \ \text{となるのは} a = 0 \text{のときにかぎる。}$$ $$|-a| = |a|$$ $$|a|^2 = a^2$$ $$|ab| = |a||b|$$ $$|\frac{a}{b}| = |\frac{a}{b}| \quad \text{ただし,} \ b \ne 0$$

$$a > 0, b > 0 \ \text{のとき}$$ $$\sqrt{ab} = \sqrt{a}\sqrt{b}$$ $$\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}}$$

$$\sqrt{a + b + 2\sqrt{ab}} = \sqrt{a} + \sqrt{b}$$ $$\sqrt{a + b - 2\sqrt{ab}} = \sqrt{a} - \sqrt{b} \quad \text{ただし,} \ a > b \ \text{とする。}$$

$$a < b \implies a + c < b + c, \quad a -c < b -c$$ $$a < b, c > 0 \implies ac < bc, \quad \frac{a}{c} < \frac{b}{c}$$ $$a < b, c < 0 \implies ac > bc, \quad \frac{a}{c} > \frac{b}{c}$$ $$\implies \ \text{は "ならば" の意味}$$

$$|x| = a \iff x = \pm a$$ $$|x| < a \iff -a < x < a$$ $$|x| > a \iff x < -a \ \text{または} \ a < x$$

$$\text{２次方程式} \ ax^2 + bx + c = 0 \ \text{の解は}$$ $$x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a}$$

$$D > 0 \iff \text{異なる２つの実数解をもつ}$$ $$D = 0 \iff \text{１つの実数解（重解）をもつ}$$ $$D < 0 \iff \text{実数解をもたない}$$

$$y = a(x -p)^2 + qのグラフは、$$ $$y = ax^2のグラフを$$ $$x軸の方向にp、y軸の方向にq$$ $$だけ平行移動した放物線である。$$ $$軸は直線y = p、頂点は点(p, q)$$

$$２次関数y = ax^2 + bx + c$$ $$のグラフとx軸の共有点のx座標は、$$ $$２次方程式$$ $$ax^2 + bx + c = 0$$ $$の実数解である。$$

$$２次方程式ax^2 + bx + c + 0が２つの解、\alpha、\betaをもつとき、$$ $$a > 0、\alpha > \betaならば$$ $$ax^2 + bx + c > 0の解はx < \alpha、\beta < x$$ $$ax^2 + bx + c < 0の解は\alpha < x < \beta$$

$$\tan A = \frac{a}{b}$$ $$a = b \tan A$$

$$\sin A = \frac{a}{c}, \cos A = \frac{b}{c}$$ $$a = c \sin A, b = c \cos A$$

$$\tan A = \frac{\sin A}{\cos A}$$ $$\sin^2 A + \cos^2 A = 1$$ $$1 + \tan^2 A = \frac{1}{\cos^2 A}$$

$$\sin(90° - A) = \cos A$$ $$\cos(90° - A) = \sin A$$ $$\tan(90° - A) = \frac{1}{\tan A}$$

$$\sin(180° - \theta) = \sin \theta$$ $$\cos(180° - \theta) = - \cos \theta$$ $$\tan(180° - \theta) = - \tan \theta$$

$$\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2R$$ $$Rは\triangle ABCの外接円の半径$$

$$a^2 = b^2 + c^2 - 2bc{\cos A}$$ $$b^2 = c^2 + a^2 - 2ca{\cos B}$$ $$c^2 = a^2 + b^2 - 2ab{\cos C}$$