Đặt:\(\dfrac{a}{b}=\dfrac{c}{d}=@\Leftrightarrow\left\{{}\begin{matrix}a=b@\\c=d@\end{matrix}\right.\)

khi đó: \(\dfrac{a^{2017}+b^{2017}}{c^{2017}+d^{2017}}=\dfrac{b^{2017}@^{2017}+b^{2017}}{d^{2017}@^{2017}+d^{2017}}=\dfrac{b^{2017}\left(@^{2017}+1\right)}{d^{2017}\left(@^{2017}+1\right)}=\dfrac{b^{2017}}{d^{2017}}\)

\(\dfrac{\left(a-b\right)^{2017}}{\left(c-d\right)^{2017}}=\dfrac{\left(b@-b\right)^{2017}}{\left(d@-d\right)^{2017}}=\dfrac{\left[b\left(@-1\right)\right]^{2017}}{\left[d\left(@-1\right)\right]^{2017}}=\dfrac{b^{2017}}{d^{2017}}\)

Ta có điều phải chứng minh

Đặt a/b=c/d=k

=>a=bk; c=dk

\(\dfrac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\dfrac{b^{2017}\cdot k^{2017}+d^{2017}\cdot k^{2017}}{b^{2017}+d^{2017}}=k^{2017}\)

\(\dfrac{\left(a+c\right)^{2017}}{\left(b+d\right)^{2017}}=\dfrac{\left(bk+dk\right)^{2017}}{\left(b+d\right)^{2017}}=k^{2017}\)

Do đó: \(\dfrac{a^{2017}+c^{2017}}{b^{2017}+d^{2017}}=\dfrac{\left(a+c\right)^{2017}}{\left(b+d\right)^{2017}}\)

Ta có:

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

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

Tới đây thì đơn giản rồi nhé.

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

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

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

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

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

từ (1) và (2) \(\Rightarrow1< A< 2\)

vay A \(\notin Z\)

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