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đáp án: a < b < c
vd:
a => 1 + 5 = 6
b => 1 + 7 = 8
c => 1 + 10 = 11
dễ thấy : a < b < c
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Vì a,b,c là các số tự nhiên khác 0 nên a,b,c > 0.
Do vậy a < a + b < a + b + c
b < b + c < a + b + c
c < c + a < a + b + c
\(\Rightarrow\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)
Ta có: a<b
\(\Rightarrow a+c< b+c\)
\(\Rightarrow a.\left(a+c\right)< b.\left(b+c\right)\)
\(\Rightarrow\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(a,b\in Z+;c\in N\right)\)
*) a + 5 = b + 7
⇒ a = b + 7 - 5
⇒ a = b + 2
⇒ a > b (1)
*) b + 7 = c + 10
⇒ b = c + 10 - 7
⇒ b = c + 3
⇒ b > c (2)
Từ (1) và (2) ⇒ a > b > c
a+5=b+7=c+10
=>a+5=b+7 và b+7=c+10
=>a-b=2>0 và b-c=10-7=3>0
=>a>b và b>c
=>a>b>c