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Ministerul Educaţiei, Cercetării şi Inovării 
Centrul Naţional pentru Curriculum şi Evaluare în Învăţământul Preuniversitar 
Soluţii 
1. −3 + 4i 
2. Se ajunge la ecuaţia ax2 + (a − 3) x − 3 = 0 , şi cum ( )2 Δ = a + 3 ≥ 0, ∀a∈∗ 
3. 2x = y; y2 − 6y + 8 = 0 ; y∈{2,4} ; x∈{1,2} 
4. ab∈{10,11,12,...,40} şi (a + b)#3⇒ 
10 
31 
p = 
5. M, N, P sunt mijloacele laturilor triunghiului, HM ⊥ BA si analoagele; HM mediatoarea[BA]si 
analoagele; H este centrul cercului circumscris +ABC 
6. 
π π= 
2 
2sin cos 
6 4 2 
BACALAUREAT 2009-MATEMATICĂ - Proba D, MT1, programa M1

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  • 1. Ministerul Educaţiei, Cercetării şi Inovării Centrul Naţional pentru Curriculum şi Evaluare în Învăţământul Preuniversitar Soluţii 1. −3 + 4i 2. Se ajunge la ecuaţia ax2 + (a − 3) x − 3 = 0 , şi cum ( )2 Δ = a + 3 ≥ 0, ∀a∈∗ 3. 2x = y; y2 − 6y + 8 = 0 ; y∈{2,4} ; x∈{1,2} 4. ab∈{10,11,12,...,40} şi (a + b)#3⇒ 10 31 p = 5. M, N, P sunt mijloacele laturilor triunghiului, HM ⊥ BA si analoagele; HM mediatoarea[BA]si analoagele; H este centrul cercului circumscris +ABC 6. π π= 2 2sin cos 6 4 2 BACALAUREAT 2009-MATEMATICĂ - Proba D, MT1, programa M1