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Bi ton. Tm tt c cc hm s f : R ! R tha mn

f(x+y)f(f(x)y) = xf(x)yf(y) (1)

vi mi s thc x; y:

Li gii. Thay x = y vo (1), ta c

f(x)f(f(x)) = xf(x)

vi mi x 2 R;t suy ra tn ti c 2 Rsao cho f(c) = 0:Thay y = f(x)vo(1) v s dng ng thc trn, ta c

f(0)f(x+f(x)) = 0

vi mi x 2 R:Thay y = xvo (1) v s dng ng thc trn, ta c

x (f(x) +f(x)) = f(0)f(x+f(x)) = 0:

T suy ra f(x) =f(x) vi mi x 6= 0: Tip tc, thay x = c vo (1) vxt y 6= 0, ta c

yf(y) = f(c+y)f(y) = f(c+y)f(y):

Do , vi mi y 6= 0 th f(y) = 0hoc f(y+c) = y: By gi, gi

A= fx2 Rjf(x) = 0g

vB =

x2 Rjf(x)6= 0 v f(x+c) = x

:

Ta xt cc trng hp sau

1. Trng hp 1. A= ;:Khi , d thy c = 0v B = R nn f(y) =y vimi y 6= 0: Do c = 0 nn f(0) = 0: Ta suy ra f(x) = x vi mi x 2 R:Th li, hm nay tha mn cc yu cu ca bi ton.

2. Trng hp 2. A 6=;:Suy ra tn ti a 6= 0 sao cho f(a) = 0:

(a) Nu B = ; th ta c A = R nn

f(x) =

0 vi x 6= 0k vi x = 0

trong k l mt s thc ty . Th li, ta thy hm ny tha mncc yu cu.

(b) Nu B 6=; th tn ti b6= 0 sao cho f(b)6= 0 v f(b+c) = b: Thayx= a v y = b vo (1) vi ch f(b) = f(b);ta c

f(a+b) = b:

T y d thy a +b6=c: Xt hai kh nng sau:

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Nuf(a+bc)6= 0 th ta c a +bc2 B nn

b= f(a+b) = f(a+bc+c) = a+bc:

Suy ra a = c: iu ny chng t A ch c duy nht mt phn tla: Do , vi mi x 6= 0; ath

f(x+a) = x:

Cho x = 2a, ta c 2a= f(a) = f(a) = 0;v l.

Mu thun trn chng tf(a +bc) = 0:V ta cng suy ra rngc6= 0(v nu c= 0th ta s c b= f(a+b) =f(a+bc) = 0cng dn n mu thun). Do vy, ta c f(c) = f(c) = 0:Thay x =cv y = a+b vo (1), ta c

(a+b)b= (a+b)f(a+b) = 0:

Suy ra b =a:Tuy nhin, lc ny ta li c

f(b) = f(a) = f(a) = 0(v l):

Vy trng hp ny khng xy ra.

Tm li, bi ton c hai nghim hm nh tm c trn.

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