A = \(\frac{2016-b-c}{2016}\)- c +\(\frac{2016-a-c}{2016}\)- a + \(\frac{2016-a-b}{2016}\) - b

= 3 - \(\frac{2\left(a+b+c\right)}{2016}\)- (a + b + c)

= 3 - 2 - 2016 = -2015

Nó là số nguyên mà bạn.

\(A=\dfrac{a}{a+b+c-c}+\dfrac{b}{a+b+c-a}+\dfrac{c}{a+b+c-b}\\ A=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\\ \Rightarrow A>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=1\left(1\right)\\ A< \dfrac{a+c}{a+b+c}+\dfrac{b+a}{a+b+c}+\dfrac{c+b}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\left(2\right)\\ \left(1\right)\left(2\right)\Rightarrow1< A< B\\ \Rightarrow A\notin Z\)

Ta có : 

Thay \(a+b+c=2016\) vào A ta có : 

\(A=\frac{a}{2016-c}+\frac{b}{2016-a}+\frac{c}{2016-b}\)

\(A=\frac{a}{a+b+c-c}+\frac{b}{a+b+c-a}+\frac{c}{a+b+c-b}\)

\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}>\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)

\(\Rightarrow\)\(A>1\)\(\left(1\right)\)

Lại có : 

\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}< \frac{a+b}{a+b+c}+\frac{b+c}{a+b+c}+\frac{c+a}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)

\(\Rightarrow\)\(A< 2\)\(\left(2\right)\)

Từ (1) và (2) suy ra : \(1< A< 2\)

Vậy A không phải là số nguyên 

Chúc bạn học tốt ~

Ta có:

\(A=\frac{a}{2016-c}+\frac{b}{2016-a}+\frac{c}{2016-b}\)

\(A=\frac{a}{a+b+c-c}+\frac{b}{a+b+c-a}+\frac{c}{a+b+c-b}\)

\(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{a+c}\)

tự làm tiếp nhé!
 

Cảm ơn hai bạn rất nhiều!!!

 a+b+c=0 => a^2+b^2+c^2+2ab+2bc+2ca = 0 => a^2+b^2+c^2=0
=> a^2+b^2+c^2 = ab+bc+ca
=> 2a^2+2b^2+2c^2 = 2ab+2bc+2ca
=> (a-b)^2 + (b-c)^2 + (c-a)^2 = 0
=> a=b=c, mà a+b+c=0 => a=b=c=0

thay vào

M=(0-2016)²⁰¹⁶⁺(0-2016)²⁰¹⁶-(0-2016)²⁰¹⁶=(-2016)²⁰¹⁶=2016²⁰¹⁶

^{Chúc bạn hoc tốt ùng hộ nha}