Tìm f: $\mathbb{R}\rightarrow \mathbb{R}$ thỏa mãn $f(x+f(y))=f(x)+\frac{xf(4y)}{8}+f(f(y)) ,\forall x,y\in \mathbb{R}$
$f(x+f(y))=f(x)+\frac{xf(4y)}{8}+f(f(y)) ,\forall x,y\in \mathbb{R}$
Started By mduc123, 13-05-2018 - 16:23
#1
Posted 13-05-2018 - 16:23
#2
Posted 16-05-2018 - 20:19
Lời giải:
P( f(x); y) $\Rightarrow f(f(x)+f(y))=f(f(x))+f(f(y))+\frac{f(x)f(4y)}{8}$
P( f(y); x) $\Rightarrow f(f(y)+f(x))=f(f(y))+f(f(x))+\frac{f(y)f(4x)}{8}$
$\Rightarrow f(x)f(4y)=f(y)f(4x)\Rightarrow \frac{f(4y)}{f(y)}=\frac{f(4x)}{f(x)}=8a$
P( -f(y); y) $\Rightarrow f(0)=0=f(-f(y))-af(y)^{2}+f(f(y))\Rightarrow f(x)+f(-x)=ax^{2}$
Thay x bởi 4x, tính theo 2 cách ta được a=2
$f(x)+f(-x)=2x^{2}\Rightarrow f(x)=x^{2}$
Thử lại:Thỏa mãn
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