Chủ đề này có 1 trả lời

[Ispectorgadget]

Tìm tất cả các hàm $f:\mathbb{R}\to \mathbb{R}$ thỏa mãn $$f(x+y)+f(x)f(y)=(1+y)f(x)+(1+x)f(y)+f(xy)$$

[perfectstrong]

Lời giải:
Kí hiệu $a:=b$ là thay $a$ bởi $b$.
\[
\begin{array}{l}
f\left( {x + y} \right) + f\left( x \right)f\left( y \right) = \left( {1 + y} \right)f\left( x \right) + \left( {1 + x} \right)f\left( y \right) + f\left( {xy} \right),\forall x,y \in R,\left( 1 \right) \\
x: = 0,y: = 0,\left( 1 \right) \Rightarrow f\left( 0 \right) + f\left( 0 \right)^2 = 3f\left( 0 \right) \Rightarrow f\left( 0 \right) = 0 \vee f\left( 0 \right) = 2 \\
\end{array}
\]
Nếu $f(0)=2$ thì\[
\begin{array}{l}
y: = 0,\left( 1 \right) \Rightarrow f\left( x \right) + 2f\left( x \right) = f\left( x \right) + 2\left( {1 + x} \right) + 2 \Rightarrow f\left( x \right) = x + 2 \\
\left( 1 \right) \Leftrightarrow - 2xy = 0:False \Rightarrow f\left( 0 \right) = 0 \\
y: = - 1,\left( 1 \right) \Rightarrow f\left( {x - 1} \right) + f\left( x \right)f\left( { - 1} \right) = \left( {1 + x} \right)f\left( { - 1} \right) + f\left( { - x} \right),\left( 2 \right) \\
x: = 1 \Rightarrow f\left( 1 \right)f\left( { - 1} \right) = 3f\left( { - 1} \right) \Rightarrow \left[ \begin{array}{l}
f\left( 1 \right) = 3 \\
f\left( { - 1} \right) = 0 \\
\end{array} \right. \\
\end{array}
\]
Nếu $f(1)=3$ thì \[
\begin{array}{l}
y: = 1,\left( 1 \right) \Rightarrow f\left( {x + 1} \right) + 3f\left( x \right) = 2f\left( x \right) + 3\left( {1 + x} \right) + f\left( x \right) \\
\Rightarrow f\left( {x + 1} \right) = 3\left( {x + 1} \right),\forall x \Rightarrow f\left( x \right) = 3x,\forall x \Rightarrow \left( 1 \right):True \\
\end{array}
\]
Nếu $f(-1)=0$ thì
\[
\begin{array}{l}
y: = - 1,\left( 1 \right) \Rightarrow f\left( {x - 1} \right) = f\left( { - x} \right),\forall x \in R,\left( 3 \right) \Rightarrow f\left( { - x - 1} \right) = f\left( x \right) \\
y: = - y - 1,\left( 1 \right) \Rightarrow f\left( {x - y - 1} \right) + f\left( x \right)f\left( { - y - 1} \right) = - yf\left( x \right) + \left( {1 + x} \right)f\left( { - y - 1} \right) + f\left( { - x\left( {y + 1} \right)} \right) \\
\Rightarrow f\left( {x - y - 1} \right) + f\left( x \right)f\left( y \right) = - yf\left( x \right) + \left( {1 + x} \right)f\left( y \right) + f\left( {xy + x - 1} \right),\left( 4 \right) \\
x: = 1 \Rightarrow f\left( { - y} \right) + f\left( 1 \right)f\left( y \right) = - yf\left( 1 \right) + 2f\left( y \right) + f\left( y \right) \\
\Rightarrow f\left( {y - 1} \right) = f\left( y \right)\left( {3 - f\left( 1 \right)} \right) - yf\left( 1 \right),\left( 5 \right) \\
\left( 1 \right),\left( 4 \right) \Rightarrow f\left( {x + y} \right) - f\left( {x - y - 1} \right) = \left( {1 + y} \right)f\left( x \right) + f\left( {xy} \right) + yf\left( x \right) - f\left( {xy + x - 1} \right) \\
x: = 1 \Rightarrow f\left( {y + 1} \right) - f\left( { - y} \right) = \left( {1 + y} \right)f\left( 1 \right) + f\left( y \right) + yf\left( 1 \right) - f\left( y \right) \\
\Rightarrow f\left( {y + 1} \right) = f\left( {y - 1} \right) + f\left( 1 \right) + 2yf\left( 1 \right) = f\left( y \right)\left( {3 - f\left( 1 \right)} \right) - yf\left( 1 \right) + f\left( 1 \right) + 2yf\left( 1 \right)\left( {do\left( 5 \right)} \right) \\
\left. \begin{array}{l}
\Rightarrow f\left( {y + 1} \right) = f\left( y \right)\left( {3 - f\left( 1 \right)} \right) + f\left( 1 \right) + yf\left( 1 \right) \\
f\left( y \right) = f\left( {y + 1} \right)\left( {3 - f\left( 1 \right)} \right) - \left( {y + 1} \right)f\left( 1 \right) \\
\end{array} \right\} \\
\Rightarrow f\left( y \right) = \left[ {f\left( y \right)\left( {3 - f\left( 1 \right)} \right) + f\left( 1 \right) + yf\left( 1 \right)} \right]\left( {3 - f\left( 1 \right)} \right) - \left( {y + 1} \right)f\left( 1 \right) \\
= f\left( y \right)\left( {3 - f\left( 1 \right)} \right)^2 + yf\left( 1 \right)\left( {3 - f\left( 1 \right)} \right) + f\left( 1 \right)\left( {3 - f\left( 1 \right)} \right) - yf\left( 1 \right) - f\left( 1 \right) \\
\Rightarrow f\left( y \right) = ay + b,\forall y \\
\left\{ \begin{array}{l}
f\left( 0 \right) = 0 \\
f\left( { - 1} \right) = 0 \\
\end{array} \right. \Rightarrow f \equiv 0 \\
\end{array}
\]
Thử lại:
Kết luận: $f \equiv 0$ hoặc $f(x)=3x$.