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Concursul interjudetean de matematica al Revistei SINUSEditia a II-a, Suceava, 18 noiembrie 2006Etapa finalaClasa a VII-a⎧1. a) Determinati elementele multimii: { } 2 x + 5⎫M = ⎨x∈ \ 2 | ∈ ⎬.⎩¢ x − 2¢⎭b) Stiind ca numerele întregi x, y, z verifica relatia xy= z 2+ z( x− y)− 5, aflati x+y .SINUS 2(5)/2006Solutie:x− 2/2x+ 5⇒x−2/2 x− 2 + 9⇒ x−2/9⇒x−2∈ ± 1, ± 3, ± 9 ………………….1pa) ( ) { }x∈{ −7, − 1,1,3,5,11}………………………………………………………………………1pb) xy= z 2+ z( x− y) −5⇔ ( x+ z)( y− z)= 5………………………………………....…2p( x z)( y z) ( 5)( 1) ( 1)( 5) ( 1)( 5) ( 5)( 1)+ − = − − = + + = − − = + + ……………………………2pFinalizare x+ y = 6……………………………………………………………………….1p
Concursul interjudetean de matematica al Revistei SINUSEditia a II-a, Suceava, 18 noiembrie 2006Etapa finalaClasa a VII-a1 1 1 6∗2. Sa se arate ca: + + ... + > ,( ∀)n ∈¥.5n+ 1 5n+2 25n5Corneliu Romascu, SuceavaSolutie:1 1 4Utilizam inegalitatea + > , ( ∀ ) ab , > 0, a ≠ ba b a + b1 1 4Avem: + >5n+ 1 25n 30n+1………………………1 1 4+ >15n 15n+ 1 30n+ 1…………………………………………………………….4p1 1 1 4+ + ... + > ⋅10n…………………………………………………..2p5n+ 1 5n+ 2 25n 30n+140n6> ⇔ 200n> 180n+6(A)………………………………………………………..1p30n+ 1 5
Page 1 and 2: Clasa a IV-a1. Suma a 3 numere este
Page 3 and 4: Concursul interjudetean de matemati
Page 7 and 8: Clasa a VI-a1. Fie unghiul ascutit
Page 9: Concursul interjudetean de matemati
Page 17 and 18: ceea ce, dupa ce trecem totul într
Page 19 and 20: PANBMC( )( )p−m b−am−n b−aM
Page 21 and 22: Solutie: a1= 1, a2= 1, a3= 8, a4= 3
Page 23: Astfel G( x+ 6) − G( x) = G( x+ 6

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