Chủ đề này có 1 trả lời

[Luffy 97]

Tìm các hàm $f:\mathbb{Z}\rightarrow \mathbb{R}$ thỏa mãn:

$f(m)+f(n)=f(mn)+f(mn+m+n)$,  $\forall m,n\in\mathbb{Z}$

 

 

[perfectstrong]

Lời giải:

\[
\begin{array}{l}
 \forall m,n \in Z:f\left( m \right) + f\left( n \right) = f\left( {mn} \right) + f\left( {mn + m + n} \right),\left( 1 \right) \\
 m: = 1,\left( 1 \right) \Rightarrow f\left( {2n + 1} \right) = f\left( 1 \right),\forall n \in Z,\left( 2 \right) \\
 m: = 2,n: = 2k + 1,\left( 1 \right) \Rightarrow f\left( 2 \right) = f\left( {4k + 2} \right),\forall k \in Z,\left( 3 \right) \\
 m: =  - 1,\left( 1 \right) \Rightarrow f\left( n \right) = f\left( { - n} \right),\forall n \in Z \\
 m: =  - 2,n: = 4,\left( 1 \right) \Rightarrow f\left( { - 2} \right) + f\left( 4 \right) = f\left( { - 8} \right) + f\left( { - 6} \right) \Rightarrow f\left( 4 \right) = f\left( { - 8} \right) = f\left( 8 \right) \\
 m = n: = 2,\left( 1 \right) \Rightarrow 2f\left( 2 \right) = f\left( 4 \right) + f\left( 8 \right) \Rightarrow f\left( 2 \right) = f\left( 4 \right) \\
 n: = 4,m: = 2k + 1,\left( 1 \right) \Rightarrow f\left( 4 \right) = f\left( {8k + 4} \right),\forall k \\
 m: =  - 2,n: =  - 4k - 2,\left( 1 \right) \Rightarrow f\left( { - 2} \right) + f\left( { - 4k - 2} \right) = f\left( {8k + 4} \right) + f\left( {4k} \right) \Rightarrow 2f\left( 2 \right) = f\left( 4 \right) + f\left( {4k} \right) \\
  \Rightarrow f\left( {4k} \right) = f\left( 2 \right),\forall k,\left( 4 \right) \\
 \left( 2 \right),\left( 3 \right),\left( 4 \right) \Rightarrow f\left( n \right) = \left\{ \begin{array}{l}
 a \text{ nếu } n \vdots 2\\
 b \text{ nếu } n \not \vdots 2\\
 \end{array} \right. \\
 \end{array}
\]
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