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Có : a/ab+a+1 = a/ab+a+abc = 1/b+1+bc = 1/bc+b+1
c/ca+c+1 = bc/abc+bc+b = b/1+bc+b = b/bc+b+1
=> A = 1+bc+b/bc+b+1 = 1
Tk mk nha
BÀI 1:
\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{a\left(bc+b+1\right)}+\frac{abc}{ab\left(ca+c+1\right)}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a} +\frac{abc}{a^2bc+abc+ab}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}\) (thay abc = 1)
\(=\frac{a+ab+1}{a+ab+1}=1\)
đặt a = 10k, b = 3k
\(\Rightarrow\frac{3\times10k-2\times3k}{10k-3\times3k}=\frac{30k-6k}{10k-9k}=\frac{24k}{k}=24\)24
Ta có: \(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\) \(\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}\)\(\Rightarrow\frac{a}{ab}+\frac{b}{ab}=\frac{b}{bc}+\frac{c}{bc}=\frac{c}{ca}+\frac{a}{ca}\)
\(\Rightarrow\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\)
+) \(\frac{1}{b}+\frac{1}{a}=\frac{1}{c}+\frac{1}{b}\) \(\Rightarrow\frac{1}{a}=\frac{1}{c}\) => a = c (1)
+) \(\frac{1}{c}+\frac{1}{b}=\frac{1}{a}+\frac{1}{c}\)\(\Rightarrow\frac{1}{b}=\frac{1}{a}\) => a = b (2)
Từ (1), (2) => a = b = c
Lại có: (a - b)3 + (b - c)3 + (c - a)3 = (a - a)3 + (b - b)3 + (c - c)3 = 03 + 03 + 03 = 0