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46 Costel ANGHEL

(a) a 7 − a 4 = 3r.
a n − a m
(b) = r.
n − m
(c) a m (n − p) + a n (p − m) + a p (m − n) = 0.

* * *

◦
◦
4. Se consider˘a triunghiul dreptunghic ABC, m(^A) = 90 , m(^B) = 30 , M mijlocul
AN −−→
laturii [BC] s , i N ∈ (AB) astfel ˆıncˆat = x. Aflat , i x s , tiind c˘a vectorii MN s , i
−−→ AB
BC sunt perpendiculari.
Vasile Ienut ,as ,, Baia Mare, Supliment G.M.-B nr. 3/2012

Clasa a X-a

1. Ar˘atat , i c˘a dac˘a log 20 = a s , i log 5 = b, atunci 5ab + a − b = 2.
50 40
Costel Anghel, Bircii

2
2
2. (a) Ar˘atat , i c˘a dac˘a a, b ∈ R, atunci a + b ≥ 2ab.
2
2
2
(b) Demonstrat , i inegalitatea a + b + c ≥ ab + bc + ac, ∀ a, b, c ∈ R.
x
x
x
x
x
x
(c) S˘a se rezolve ecuat , ia 4 + 25 + 49 = 10 + 14 + 35 .
Costel Anghel, Bircii
2
3. Fie α ∈ C o r˘ad˘acin˘a a ecuat , iei z + z + 1 = 0.
(a) Calculat , i α 2013 .
(b) Determinat , i o ecuat , ie de gradul al II-lea cu coeficient , ii reali care are r˘ad˘acina
complex˘a z = 3 − i.
2
4
(c) Rezolvat , i ˆın C ecuat , ia z + 2z − 3 = 0.
Florea Badea, Scornices , ti

4. S˘a se demonstreze egalitatea de mai jos, pentru n > 2l:
l−1 2l
X 2 2k X k
(n − 2k) A = A
n n
k=0 k=1
.

Octavian Marinescu-Ghemeci