初中數學（香港課程）/恆等式與公式/2

平方差

證明 $(a + b) (a - b) \equiv a^{2} - b^{2}$ ：

$\begin{matrix} L.H.S. & = (a + b) (a - b) \\ = a^{2} - a b + b a - b^{2} \\ = a^{2} - b^{2} \\ = R.H.S. \\ ∴ (a + b) (a - b) \equiv a^{2} - b^{2} \end{matrix}$

上述恆等式稱作平方差，記為

(a + b) (a - b) \equiv a^{2} - b^{2}

示範例子1

利用恆等式，展開

$(9 + m) (m - 9)$
$(4 c + 7 d) (7 d - 4 c)$
$(8 - 5 k) (- 5 k - 8)$

解

$\begin{matrix} (9 + m) (m - 9) & = (m + 9) (m - 9) \\ = m^{2} - 9^{2} \\ = m^{2} - 81 \end{matrix}$
$\begin{matrix} (4 c + 7 d) (7 d - 4 c) & = (7 d)^{2} - (4 c)^{2} \\ = 7 d \cdot 7 d - 4 c \cdot 4 c \\ = 49 d^{2} - 16 c^{2} \end{matrix}$
$\begin{matrix} (8 - 5 k) (- 5 k - 8) & = (- 5 k + 8) (- 5 k - 8) \\ = (- 5 k)^{2} - 8^{2} \\ = 25 k^{2} - 64 \end{matrix}$

同類習題1

{{{3}}}

示範例子2

展開

$6 (2 h + 1) (2 h - 1)$
$(r + 3) (r - 3) (r^{2} + 9)$
解析失败 (语法错误): {\displaystyle -5(w^2+6)(w^2-6) |sol= # <math>\begin{align} 6(2h+1)(2h-1)&=6(2h\cdot 2h-1)\\ &=6(4h^2-1)=24h \end{align}}
$\begin{matrix} (r + 3) (r - 3) (r^{2} + 9) & = (r^{2} - 9) (r^{2} + 9) \\ = (r^{2})^{2} - 9^{2} \\ = r^{4} - 8 \end{matrix}$
$\begin{matrix} - 5 (w^{2} + 6) (w^{2} - 6) & = - 5 (w^{2} \cdot w^{2} - 6^{2}) \\ = - 5 (w^{4} - 36) \\ = 180 - 5 w^{4} \end{matrix}$

同類習題2

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完全平方

證明 $(a + b)^{2} \equiv a^{2} + 2 a b + b^{2}$ :

$\begin{matrix} L.H.S. & = (a + b)^{2} \\ = (a + b) (a + b) \\ = a^{2} + a b + b a + b^{2} \\ = a^{2} + 2 a b + b^{2} \\ = R.H.S. \\ ∴ (a + b)^{2} \equiv a^{2} + 2 a b + b^{2} \end{matrix}$

上述恆等式稱作和平方，記作

(a + b)^{2} \equiv a^{2} + 2 a b + b^{2}

只要b變為 $- b$ ，變為差平方，記作

(a - b)^{2} \equiv a^{2} - 2 a b + b^{2}

兩個恆等式均為完全平方

示範例子3

利用恆等式，展開

$(z - 3)^{2}$
$(4 u + 7 v)^{2}$
$(1 - 8 t)^{2}$

解

$\begin{matrix} (z - 3)^{2} & = z^{2} - 2 \cdot 3 z + 3^{2} \\ = z^{2} - 6 z + 9 \end{matrix}$
$\begin{matrix} (4 u + 7 v) & = (4 u)^{2} - 2 \cdot 4 u \cdot 7 v + (7 v)^{2} \\ = 4 u \cdot 4 u - 56 u v + 7 v \cdot 7 v \\ = 16 u^{2} - 56 u v + 49 v^{2} \end{matrix}$
$\begin{matrix} (1 - 8 t)^{2} & = 1^{2} - 2 \cdot 1 \cdot 8 t + (8 t)^{2} \\ = 1 - 16 t + 64 t^{2} \end{matrix}$

同類習題3

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示範例子4

$(\frac{p}{9} + \frac{q}{6})^{2}$
$(c - 7)^{2} + (8 c + 1)^{2}$

解

$\begin{matrix} (\frac{p}{9} + \frac{q}{6})^{2} & = (\frac{p}{9})^{2} + 2 \cdot \frac{p}{9} \cdot \frac{q}{6} + (\frac{q}{6})^{2} \\ = \frac{p}{9} \cdot \frac{p}{9} + 2 \cdot \frac{p q}{54} + \frac{q}{6} \cdot \frac{q}{6} \\ = \frac{p^{2}}{81} + \frac{p q}{27} + \frac{q^{2}}{36} \end{matrix}$

$\begin{matrix} (c - 7)^{2} + (8 c + 1)^{2} & = c^{2} - 14 c + 49 + 64 c^{2} + 16 c + 1 \\ = 65 c^{2} + 2 c + 50 \end{matrix}$

同類習題4

{{{3}}}