\(a+b+c=2017\Rightarrow A=\dfrac{a}{a+b+c-c}+\dfrac{b}{a+b+c-b}+\dfrac{c}{a+b+c-a}\)

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

\(\Rightarrow A< 2\left(1\right)\)

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

\(\Rightarrow A>1\left(2\right)\)

từ (1) và (2) \(\Rightarrow1< A< 2\)

vay A \(\notin Z\)

\(a^2+c^2=b^2+d^2\Leftrightarrow2\left(a^2+c^2\right)=a^2+b^2+c^2+d^2\)

\(2\left(a^2+c^2\right)⋮2\Rightarrow a^2+b^2+c^2+d^2⋮2\)

Xét: \(\left(a^2+b^2+c^2+d^2\right)-\left(a+b+c+d\right)=a\left(a-1\right)+b\left(b-1\right)+c\left(c-1\right)+d\left(d-1\right)⋮2\) (tích 2 số nguyên liên tiếp thì chia hết cho 2)

\(\Rightarrow a+b+c+d⋮2;a+b+c+d>2\left(a;b;c;d\in N>0\right)\)

\(\Rightarrow a+b+c+d\) là hợp số (đpcm)

b^2=ac

b^2+2017bc=ac+2017bc

b(b+2017c)=c(a+2017b)

b/c=(a+2017b)/(b+2017c)

(b/c)^2=((a+2017b)/(b+2017c))^2

b^2/c^2=(a+2017b)^2/(b+2017c)^2

thế b^2=ac ta có 

ac/c^2=(a+2017b)^2/(b+2017c)^2 

a/c=(a+2017b)^2/(b+2017c)^2 

Gọi $d$ là ước số (chung) của $a,b$

Đặt \(\left\{\begin{matrix} a=md\\ b=nd\end{matrix}\right.(m,n\in\mathbb{Z}^+)\)

Ta có:

\(\frac{a+1}{b}+\frac{b+1}{a}\in\mathbb{Z}\)

\(\Leftrightarrow \frac{a^2+b^2+a+b}{ab}\in\mathbb{Z}\)

\(\Leftrightarrow \frac{(a+b)^2+(a+b)-2ab}{ab}\in\mathbb{Z}\Leftrightarrow \frac{(a+b)^2+(a+b)}{ab}\in\mathbb{Z}\)

\(\Leftrightarrow (a+b)^2+a+b\vdots ab\)

\(\Leftrightarrow (md+nd)^2+md+nd\vdots mnd^2\)

\(\Leftrightarrow d(m+n)^2+m+n\vdots mnd\)

\(\Rightarrow d(m+n)^2+m+n\vdots d\Rightarrow m+n\vdots d\)

Mà \(m+n\neq 0\). Do đó suy ra \(m+n\geq d\)

\(\Rightarrow d(m+n)\geq d^2\) hay \(a+b\geq d^2\Rightarrow d\leq \sqrt{a+b}\)

Ta có đpcm.

Tương tụ bài trên

A,B,C,E đạt giá trị nhỏ nhất =0

a)x=5

b)x=-5

c)x=2

d)x=-1