$\text{C1}$
Đặt $\frac{a}{a+b}$ + $\frac{b}{b+c}$ + $\frac{c}{c+a}$ là M
Ta có: M=$\frac{a}{a+b}$ + $\frac{b}{b+c}$ + $\frac{c}{c+a}$
⇒$\frac{2a}{a+b+c}$ < $\frac{a}{a+b}$ < $\frac{a+b}{a+b}$
⇒$\frac{2b}{a+b+c}$ < $\frac{b}{b+c}$ < $\frac{b+c}{b+c}$
⇒$\frac{2c}{a+b+c}$ < $\frac{c}{c+a}$ < $\frac{c+a}{c+a}$
Ta có : $\frac{2a}{a+b+c}$ + $\frac{2b}{a+b+c}$ + $\frac{2c}{a+b+c}$ = $\frac{2(a+b+c)}{a+b+c}$ =2
$\frac{a+b}{a+b}$ + $\frac{b+c}{b+c}$ + $\frac{c+a}{c+a}$=1+1+1=3
⇒2<M<3
Vậy $\frac{a}{a+b}$ + $\frac{b}{b+c}$ + $\frac{c}{c+a}$ ko phải là số nguyên
$\text{C2}$
Đặt $\frac{a}{a+b}$ + $\frac{b}{b+c}$ + $\frac{c}{c+a}$ là A
Ta có: $\frac{a}{a+b}$ + $\frac{b}{b+c}$ + $\frac{c}{c+a}$ > $\frac{a}{a+b+c}$ + $\frac{b}{a+b+c}$ + $\frac{c}{a+b+c}$=$\frac{a+b+c}{a+b+c}$ =1
Ta có: $\frac{a}{a+b}$ + $\frac{b}{b+c}$ + $\frac{c}{c+a}$<$\frac{2a}{a+b+c}$ + $\frac{2b}{a+b+c}$ + $\frac{2c}{a+b+c}$=$\frac{2(a+b+c)}{a+b+c}$ = 2
⇒1<A<2
Vậy $\frac{a}{a+b}$ + $\frac{b}{b+c}$ + $\frac{c}{c+a}$ ko phải là số nguyên
THEO 2 CÁCH NHÉ
NHỚ CHO CTLHN NHÉ
