8/16/2019 Gradska Takmicenja u Hrvatskoj 1992 Do 2006

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1992 −2006

a 2 , b2 , c2 1

a + b +

1b + c

= 2a + c

n 2 −n + 2n

3

+ 2 n2

−n + 1

n∈

N

a,b,c

a 2 −bca (1 −bc)

= b2 −acb(1 −ac )

, abc (1 −bc)(1 −ac ) ̸= 0 .

a̸=

b a + b + c = 1a

+ 1b

+ 1c

x,y,z (x,y,z ) x −y = y −z = 96

a

|3

−2

|x

||=

−34 a

a,b,c

2b + c −a

+ 2

c + a −b +

2a + b−c

> 2a

+ 2b

+ 2c

.

p∈

R

5x5x 2 + px + 45

+ x + 10x 2 + 5 x

= 2x

x + y + z = a x2 + y2 + z 2 = b2 x − 1 + y− 1 + z− 1 = c− 1 x 3 + y3 + z3

0, 1, . . . , 9 13 14 15

a,b,c,d,e,f

(a −b)2 + ( b −c)

2 + ( c −d)2 + ( d −e)

2 + ( e −f )2 + ( f −a )

2

≥18.

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|x |+ |y|+ |x + y| ≤2? a ̸= 0 b, c

d a

b + c + d

0.abc

a b

a 3 −3ab2 = 44 , b3 −3a

2 b = 8 .

a 2 + b2

x + y + z = 0 x7 + y7 + z7

xyz (x 4 + y4 + z4 ).

x 14x + 5

9

17x

−5

12

a a5 −a3 + a = 2

3 < a 6 < 4

4x + y + 4 √ xy −28√ x −14√ y + 48 = 0

a > 0 f (x ) = x 5 + ax x > 0

4xyz −x4

−y4

−z4 = 1

(x,y,z )

a,b,c 1

a +

1b

+ 1c

= 1 , (a −1)(b −1)(c −1) ≥8

√ 1 + 112 + 122 + √ 1 + 122 + 132 + . . . +√ 1 + 120042 + 120052 = 2005− 12005 . n 5n −23n −7

x, y x2 + xy + y2 = 4 x4 + x 2 y2 + y4 = 8 x 6 + x3 y3 + y6

A∈

N n B A A + B = 10 n

A B n = 4

n A B 10

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