Page 10 - RMGO 4
P. 10

10                                                            Probleme propuse





                                            Clasa a XII-a



                                                                        ∗
            MGO 156. Fie p un num˘ar prim de forma p = 4k + 3, k ∈ N . Demonstrat , i c˘a
                                2
                       2
            ecuat , ia  x + k + 4  = 9 nu are solut , ii ˆın Z p .
                           b
                                     b
                              b
                                      Stelian Corneliu Andronescu s , i Costel B˘alc˘au, Pites , ti
                                                        n
            MGO 157. Fie n ∈ N, n ≥ 3. Rezolvat , i ˆın R sistemul
                                                        +
                                        
                                            5        2
                                         x + 3 = x + 3x 3
                                            1
                                                     2
                                        
                                        
                                            5        2
                                        
                                           x + 3 = x + 3x 4
                                            2        3       .
                                         . . .
                                        
                                        
                                           5        2
                                           x + 3 = x + 3x 2
                                                     1
                                            n
                                                                    Mih´aly Bencze, Bras , ov
                                          ∗
                                  ∗
            MGO 158. Fie n ∈ N , a ∈ R s , i b > 0. Calculat , i integrala
                                  Z   2n−1       2n−1
                                     x    (x + a)
                                                      dx, x ∈ (0, ∞).
                                     (x + a) 4n  + bx 4n
                                                                      Daniel Jinga, Pites , ti
            MGO 159. Fie f : [−1, 1] → R o funct , ie de dou˘a ori derivabil˘a, cu derivata a
            doua continu˘a, astfel ˆıncˆat f(−1) = f(1) = 0. Demonstrat , i c˘a are loc inegalitatea
                                  1  Z  1       2
                                                               2
                                           00
                                    ·     f (x)   dx ≥ max f (x).
                                  6   −1               x∈[−1,1]
                                                                   Florin St˘anescu, G˘aes , ti
            MGO 160. Fie a, b, c, d ≥ 0 astfel ˆıncˆat ab + ac + ad + bc + bd + cd = 6.
                Demonstrat , i c˘a
                                         √
                                      2 +  2                                 √
                    2
                                  2
                             2
                         2
                   a + b + c + d +           · (abc + abd + acd + bcd) ≥ 2 4 +  2 .
                                        2
                Cˆand are loc egalitatea?
                                            Leonard Mihai Giugiuc, Drobeta Turnu Severin
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