初中数学（香港课程）/恒等式与公式/2

平方差

证明$(a+b)(a-b)\equiv a^{2}-b^{2}$：

${\begin{aligned}{\text{L.H.S.}}&=(a+b)(a-b)\\&=a^{2}-ab+ba-b^{2}\\&=a^{2}-b^{2}\\&={\text{R.H.S.}}\\\therefore (a+b)(a-b)\equiv a^{2}-b^{2}\end{aligned}}$

上述恒等式称作平方差，记为

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

示范例子1 利用恒等式，展开 $(9+m)(m-9)$ $(4c+7d)(7d-4c)$ $(8-5k)(-5k-8)$ 解 ${\begin{aligned}(9+m)(m-9)&=({\color {red}m}+{\color {blue}9})({\color {red}m}-{\color {blue}9})\\&={\color {red}m}^{2}-{\color {blue}9}^{2}\\&=m^{2}-81\end{aligned}}$ ${\begin{aligned}(4c+7d)(7d-4c)&=(7d)^{2}-(4c)^{2}\\&=7d\cdot 7d-4c\cdot 4c\\&=49d^{2}-16c^{2}\end{aligned}}$ ${\begin{aligned}(8-5k)(-5k-8)&=(-5k+8)(-5k-8)\\&=(-5k)^{2}-8^{2}\\&=25k^{2}-64\end{aligned}}$

同类习题1 {{{3}}}

示范例子2 展开 $6(2h+1)(2h-1)$ $(r+3)(r-3)(r^{2}+9)$ 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}} ${\begin{aligned}(r+3)(r-3)(r^{2}+9)&=(r^{2}-9)(r^{2}+9)\\&=(r^{2})^{2}-9^{2}\\&=r^{4}-8\end{aligned}}$ ${\begin{aligned}-5(w^{2}+6)(w^{2}-6)&=-5(w^{2}\cdot w^{2}-6^{2})\\&=-5(w^{4}-36)\\&=180-5w^{4}\end{aligned}}$

同类习题2 {{{3}}}

完全平方

证明$(a+b)^{2}\equiv a^{2}+2ab+b^{2}$:

${\begin{aligned}{\text{L.H.S.}}&=(a+b)^{2}\\&=(a+b)(a+b)\\&=a^{2}+ab+ba+b^{2}\\&=a^{2}+2ab+b^{2}\\&={\text{R.H.S.}}\\\therefore (a+b)^{2}\equiv a^{2}+2ab+b^{2}\end{aligned}}$

上述恒等式称作和平方，记作

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

只要b变为$-b$，变为差平方，记作

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

两个恒等式均为完全平方

示范例子3 利用恒等式，展开 $(z-3)^{2}$ $(4u+7v)^{2}$ $(1-8t)^{2}$ 解 ${\begin{aligned}(z-3)^{2}&=z^{2}-2\cdot 3z+3^{2}\\&=z^{2}-6z+9\end{aligned}}$ ${\begin{aligned}(4u+7v)&=(4u)^{2}-2\cdot 4u\cdot 7v+(7v)^{2}\\&=4u\cdot 4u-56uv+7v\cdot 7v\\&=16u^{2}-56uv+49v^{2}\end{aligned}}$ ${\begin{aligned}(1-8t)^{2}&=1^{2}-2\cdot 1\cdot 8t+(8t)^{2}\\&=1-16t+64t^{2}\end{aligned}}$

同类习题3 {{{3}}}

示范例子4 $({\frac {p}{9}}+{\frac {q}{6}})^{2}$ $(c-7)^{2}+(8c+1)^{2}$ 解 ${\begin{aligned}({\frac {p}{9}}+{\frac {q}{6}})^{2}&=({\frac {p}{9}})^{2}+2\cdot {\frac {p}{9}}\cdot {\frac {q}{6}}+({\frac {q}{6}})^{2}\\&={\frac {p}{9}}\cdot {\frac {p}{9}}+2\cdot {\frac {pq}{54}}+{\frac {q}{6}}\cdot {\frac {q}{6}}\\&={\frac {p^{2}}{81}}+{\frac {pq}{27}}+{\frac {q^{2}}{36}}\end{aligned}}$ ${\begin{aligned}(c-7)^{2}+(8c+1)^{2}&=c^{2}-14c+49+64c^{2}+16c+1\\&=65c^{2}+2c+50\end{aligned}}$

同类习题4 {{{3}}}