Chủ đề này có 1 trả lời

[Minhnksc]

Tìm các hàm $f:\mathbb{Q}^2\rightarrow \mathbb{Q}$ thỏa mãn

$f(x,y)+f(y,z)+f(z,x)=f(0,x+y+z)$ với mọi $x;y;z\in \mathbb{Q}$

[perfectstrong]

Lời giải vắn tắt:

\[f:{Q^2} \to Q:f\left( {x,y} \right) + f\left( {y,z} \right) + f\left( {z,x} \right) = f\left( {0,x + y + z} \right),\left( 1 \right)\]

Quy ước $a:=b$ nghĩa là "thay $a$ bởi $b$.

\[\begin{array}{c}
x = y = z: = 0,\left( 1 \right) \Rightarrow f\left( {0,0} \right) = 0\\
y = z: = 0,\left( 1 \right) \Rightarrow f\left( {x,0} \right) + f\left( {0,x} \right) = f\left( {0,x} \right) \Rightarrow f\left( {x,0} \right) = 0\forall x\left( 2 \right)\\
z: = 0,\left( 1 \right),\left( 2 \right) \Rightarrow f\left( {x,y} \right) + f\left( {0,x} \right) = f\left( {0,x + y} \right),\left( 3 \right)
\end{array}\]

 

\[\left. \begin{array}{c}
f\left( {0,x + \left( {y + z} \right)} \right)\mathop  = \limits^{\left( 3 \right)} f\left( {x,y + z} \right) + f\left( {0,x} \right)\\
f\left( {0,\left( {x + y} \right) + z} \right)\mathop  = \limits^{\left( 3 \right)} f\left( {x + y,z} \right) + f\left( {0,x + y} \right)\mathop  = \limits^{\left( 3 \right)} f\left( {x + y,z} \right) + f\left( {x,y} \right) + f\left( {0,x} \right)
\end{array} \right\} \\ \Rightarrow f\left( {x,y + z} \right) = f\left( {x + y,z} \right) + f\left( {x,y} \right)\forall x,y,z,\left( 4 \right)\]

 

\[\begin{array}{c} z: = y,\left( 4 \right) \Rightarrow f\left( {x,2y} \right) = f\left( {x + y,y} \right) + f\left( {x,y} \right)\\ z: = 2y,\left( 4 \right) \Rightarrow f\left( {x,3y} \right) = f\left( {x + y,2y} \right) + f\left( {x,y} \right) = f\left( {x + 2y,y} \right) + f\left( {x,y} \right) + f\left( {x,y} \right) = f\left( {x + 2y,y} \right) + 2f\left( {x,y} \right)\\  \Rightarrow f\left( {x,\left( {n + 1} \right)y} \right) = f\left( {x + ny,y} \right) + nf\left( {x,y} \right)\forall n \in N,\left( 5 \right) \end{array}\]

 

\[\begin{array}{l} \begin{array}{ccc}
f\left( {x,4y} \right)& \mathop = \limits^{\left( 4 \right):z: = 3y}& f\left( {x + 3y,y} \right) + f\left( {x,3y} \right)\\ & \mathop = \limits^{\left( 4 \right)}& f\left( {x + 3y,y} \right) + f\left( {x + 2y,y} \right) + 2f\left( {x,y} \right),\left( i \right)\\
f\left( {x,4y} \right)& \mathop = \limits^{\left( 4 \right):\left( {y;z} \right):= \left( {2y;2y} \right)}& f\left( {x + 2y,2y} \right) + f\left( {x,2y} \right)\\ & \mathop = \limits^{\left( 4 \right)}& \left( {f\left( {x + 2y + y,y} \right) + f\left( {x,y} \right)} \right) + \left( {f\left( {x + y,y} \right) + f\left( {x,y} \right)} \right)\\ & = & f\left( {x + 3y,y} \right) + f\left( {x + y,y} \right) + 2f\left( {x,y} \right),\left( {ii} \right) \end{array}\\
\left( i \right),\left( {ii} \right) \Rightarrow f\left( {x + 2y,y} \right) = f\left( {x + y,y} \right)\mathop  \Rightarrow \limits^{x: = x - y} f\left( {x + y,y} \right) = f\left( {x,y} \right)\forall x,y\left( 6 \right)
\end{array}\]

 

\[\begin{array}{c}
\left. \begin{array}{l}
\left( 6 \right) \Rightarrow f\left( {x + ny,y} \right) = f\left( {x,y} \right)\forall n \in N\\
x: = x - y,\left( 6 \right) \Rightarrow f\left( {x,y} \right) = f\left( {x - y,y} \right) \Rightarrow f\left( {x - ny,y} \right) = f\left( {x,y} \right)\forall n \in N
\end{array} \right\}\\
 \Rightarrow f\left( {x + ny,y} \right) = f\left( {x,y} \right)\forall n \in Z\left( {6 + } \right)
\end{array}\]

 

\[\begin{array}{c}
\left. \begin{array}{l}
\left( 5 \right),\left( 6 \right) \Rightarrow f\left( {x,ny} \right) = nf\left( {x,y} \right)\forall n \in N\\
z: = y,y: =  - y,\left( 4 \right) \Rightarrow f\left( {x,0} \right) = f\left( {x - y,y} \right) + f\left( {x, - y} \right) \Rightarrow f\left( {x, - y} \right) =  - f\left( {x - y,y} \right)\mathop  = \limits^{\left( {6 + } \right)}  - f\left( {x,y} \right)
\end{array} \right\}\\
 \Rightarrow f\left( {x,ny} \right) = nf\left( {x,y} \right)\forall n \in Z\\
y: = \frac{y}{n} \Rightarrow f\left( {x,y} \right) = nf\left( {x,\frac{y}{n}} \right)\forall n \in Z\\
f\left( {x,\frac{p}{q}y} \right) = pf\left( {x,\frac{y}{q}} \right) = \frac{p}{q}f\left( {x,y} \right) \Rightarrow f\left( {x,zy} \right) = zf\left( {x,y} \right)\forall z,\left( a \right) \Rightarrow f\left( {x,y} \right) = yf\left( {x,1} \right)\forall y,\left( b \right)
\end{array}\]

 

\[\begin{array}{c}
\left( 4 \right),\left( 6 \right) \Rightarrow f\left( {x,y + z} \right) = f\left( {x,z} \right) + f\left( {x,y} \right)\forall x,y,z,\left( {4 + } \right)\\
y: = \frac{y}{m},\left( {6 + } \right) \Rightarrow f\left( {x + \frac{n}{m}y,\frac{y}{m}} \right) = f\left( {x,\frac{y}{m}} \right)\mathop  \Rightarrow \limits^{\left( a \right)} \frac{1}{m}f\left( {x + \frac{n}{m}y,y} \right) = \frac{1}{m}f\left( {x,y} \right) \Rightarrow f\left( {x + zy,y} \right) = f\left( {x,y} \right)\forall z\\
x: = 0,z \in Q \Rightarrow f\left( {zy,y} \right) = f\left( {0,y} \right)\mathop  = \limits^{\left( b \right)} yf\left( {0,1} \right)\forall y,z \Rightarrow f\left( {x,y} \right) = ay\forall x,y
\end{array}\]