Giải:

Theo đề bài ta có:

\(\left\{\begin{matrix}2014a+3b+1\\2014^a+2014a+b\end{matrix}\right.\) là hai số lẻ

Nếu \(a\ne0\Rightarrow2014^a+2014a\) là số chẵn

Để \(2014^a+2014a+b\) là số lẻ \(\Rightarrow b\) phải là số lẻ

Nếu \(b\) là số lẻ \(\Rightarrow3b+1\) là số chẵn, do đó:

\(2014a+3b+1\) là số chẵn (không thỏa mãn)

Vậy \(a=0\)

Với \(a=0\Rightarrow\left(3b+1\right)\left(b+1\right)=225\)

Vì \(b\in N\)

\(\Rightarrow\left(3b+1\right)\left(b+1\right)=3.75=5.45=9.25=1.225\)

\(3b+1⋮̸\)\(3;3b+1>b+1\)

\(\Rightarrow\left\{\begin{matrix}3b+1=25\\b+1=9\end{matrix}\right.\)\(\Rightarrow b=8\)

Vậy: \(\left\{\begin{matrix}a=0\\b=8\end{matrix}\right.\)

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\(P=\frac{2014a}{ab+2014a+2014}+\frac{b}{bc+b+2014}+\frac{c}{ac+c+1}\)

\(P=\frac{a^2bc}{ab+a^2bc+abc}+\frac{ab}{abc+ab+a^2bc}+\frac{c}{ac+c+1}\)

\(P=\frac{ac}{1+ac+c}+\frac{1}{c+1+ac}+\frac{c}{ac+c+1}\)

\(P=\frac{ac+1+c}{1+ac+c}=1\)

\(\frac{a}{b}=\frac{c}{d}=>ad=bc=>\frac{a}{c}=\frac{b}{d}=>\frac{2014.a}{2014c}=\frac{2015b}{2015d}\)

Áp dụng dãy tỉ số bằng nhau ta có:

\(\frac{2014a}{2014c}=\frac{2015b}{2015d}=\frac{2014a-2015b}{2014c-2015d}=\frac{2014a+2015b}{2014c+2015d}\)

=>\(\frac{2014a-2015b}{2014c-2015d}=\frac{2014a+2015b}{2014c+2015d}\)

=> (2014a-2015b).(2014c+2015d)=(2014c-2015d).(2014a+2015b)

=>\(\frac{2014a-2015b}{2014a+2015b}=\frac{2014c-2015d}{2014c+2015d}\)

\(\frac{2014a}{ab+2014a+2014}+\frac{b}{bc+b+2014}+\frac{c}{ac+c+1}\)

\(=\frac{abc.a}{ab+abca+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)

\(=\frac{ac}{1+ac+c}+\frac{1}{c+1+ac}+\frac{c}{ac+c+1}=1\left(ĐPCM\right)\)

Từ \(\dfrac{ab}{2014}=\dfrac{1}{c}\Rightarrow abc=2014\) thay vào \(A\) ta có:

\(A=\dfrac{abc\cdot a}{ab+abc\cdot a+abc}+\dfrac{b}{bc+b+abc}+\dfrac{c}{ac+c+1}\)

\(=\dfrac{a^2bc}{ab+a^2bc+abc}+\dfrac{b}{b\left(ac+c+1\right)}+\dfrac{c}{ac+c+1}\)

\(=\dfrac{ac\cdot ab}{ab\left(ac+c+1\right)}+\dfrac{1}{ac+c+1}+\dfrac{c}{ac+c+1}\)

\(=\dfrac{ac}{ac+c+1}+\dfrac{1}{ac+c+1}+\dfrac{c}{ac+c+1}\)

\(=\dfrac{ac+c+1}{ac+c+1}=1\Rightarrow A=1\)