文章目录
比例恒等式(分式恒等式)
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设 a b = c d \frac{a}{b}=\frac{c}{d} ba=dc
(0)令这个比值为 k k k,则 a = k b a=kb a=kb(0-1), c = k d c=kd c=kd(0-2),以下恒等式在表达式有意义的情形下成立(例如分母不为0) -
合比定理: a + b b = c + d d \frac{a+b}{b}=\frac{c+d}{d} ba+b=dc+d
(1)- 对式(0)两边同时加 1 1 1,得 a b + 1 = c d + 1 \frac{a}{b}+1=\frac{c}{d}+1 ba+1=dc+1,通分得式(1)
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分比定理: a − b b = c − d d \frac{a-b}{b}=\frac{c-d}{d} ba−b=dc−d
(2)- 对式(0)两边同时减1,得式(2)
- 也可以由合比定理将 b b b用 − b -b −b代替得到
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合分比定理: a + b c − b = c + d c − d \frac{a+b}{c-b}=\frac{c+d}{c-d} c−ba+b=c−dc+d
(3)- 由式(1)比去式(2),即得(3)
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a c \frac{a}{c} ca= b d \frac{b}{d} db
(4)- 将(0-1,0-2)得 a c \frac{a}{c} ca= k b k d \frac{kb}{kd} kdkb= b d \frac{b}{d} db
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若 a + c b + d = k \frac{a+c}{b+d}=k b+da+c=k,即 a b = c d \frac{a}{b}=\frac{c}{d} ba=dc= a + c b + d \frac{a+c}{b+d} b+da+c= k k k
(5)- 由(0-1,0-2),得 a + c b + d \frac{a+c}{b+d} b+da+c= k ( b + d ) b + d \frac{k(b+d)}{b+d} b+dk(b+d)= k k k
分式等式链
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推广:若 a 1 b 1 \frac{a_1}{b_1} b1a1= ⋯ \cdots ⋯= a n b n \frac{a_n}{b_n} bnan= k k k,则 ∑ i = 1 n a i ∑ i = 1 n b i \frac{\sum_{i=1}^{n}a_{i}}{\sum_{i=1}^{n}b_{i}} ∑i=1nbi∑i=1nai= k k k
(6)-
设 I = { 1 , 2 , ⋯ , n } I=\set{1,2,\cdots,n} I={1,2,⋯,n}, S S S是从 I I I中任意选出 m m m个元素构成的几何 ( m ∈ [ 1 , n ] , m ∈ N + ) (m\in{[1,n]},m\in\mathbb{N_{+}}) (m∈[1,n],m∈N+),都有 ∑ i ∈ S a i ∑ i ∈ S b i \Large{\frac{\sum_{i\in S}a_{i}}{\sum_{i\in{S}}b_{i}}} ∑i∈Sbi∑i∈Sai= k k k
(6-1) -
∑ i ∈ S k i a i ∑ i ∈ S k i b i \Large{\frac{\sum_{i\in S}k_{i}a_{i}}{\sum_{i\in{S}}k_{i}b_{i}}} ∑i∈Skibi∑i∈Skiai= k k k,
(6-2)其中 k i ∈ { − 1 , 1 } k_i\in\set{-1,1} ki∈{−1,1}- 因为 a i b i = − a i − b i \frac{a_{i}}{b_{i}}=\frac{-a_{i}}{-b_{i}} biai=−bi−ai= k k k,再由结论(5),可知结论(6-2)成立
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例
- 设 y x = y + z x + z \frac{y}{x}=\frac{y+z}{x+z} xy=x+zy+z= k k k,则 k = 1 k=1 k=1
- 由性质(5), y x \frac{y}{x} xy= y + z − y x + z − x \frac{y+z-y}{x+z-x} x+z−xy+z−y= z z \frac{z}{z} zz=1;所以 k = 1 k=1 k=1,即 x = y x=y x=y
- 方法2: y = k x y=kx y=kx; y + z = k x + k z y+z=kx+kz y+z=kx+kz,联立得 k = 1 k=1 k=1,即 x = y x=y x=y