Page 93 - RMGO 3

Probleme propuse 93

MGO 107. Fie a > 0, a 6= 1. Rezolvat , i ˆın mult , imea numerelor reale ecuat , ia
2
log (x − 2a − 1) = log 2a+1 (2x + 2a − 1).
a
Marin Chirciu, Pites , ti

MGO 108. Fie numerele reale a, b s , i c astfel ˆıncˆat

th (a)th (b) + th (b)th (c) + th (c)th (a) = 1

x
e − e −x
(unde th (x) = reprezint˘a tangenta hiperbolic˘a a num˘arului real x).
x
e + e −x
a) Demonstrat , i c˘a a, b, c < 0 sau a, b, c > 0.

1 2 + th (a) + th (b) + th (c) + th (a)th (b)th (c)
b) Ar˘atat , i c˘a a + b + c = ln .
2 2 − th (a) + th (b) + th (c) + th (a)th (b)th (c)
Michel Bataille, Frant , a s , i Leonard Mihai Giugiuc, Romˆania
∗
MGO 109. Fie z 1 , z 2 , z 3 ∈ C astfelˆıncˆat |z 1 +z 2 | = |z 1 |+|z 2 |, |z 1 +z 3 | = |z 1 |+|z 3 |
s , i z 2 + z 3 = 2z 1 . Determinat , i mult , imea tuturor valorilor posibile pentru num˘arul
z 2 · z 2 + z 3 · z 3
.
z 1 · z 1
Mihai Florea Dumitrescu, Potcoava

MGO 110. Rezolvat , i ˆın mult , imea numerelor reale ecuat , ia
2
2
2
2
cos x + cos 2x · sin 3x + sin 4x = 1.
Daniel Jinga, Pites , ti

Clasa a XI-a

6
MGO 111. Fie A ∈ M n (C) astfel ˆıncˆat A = A + I n . Demonstrat , i c˘a matricea
2
A + A + I n este inversabil˘a.
Cristinel Mortici, Viforˆata

∗
MGO 112. Fie a, b ∈ C astfel ˆıncˆat a 6= ±b s , i A, B ∈ M 4 (C) astfel ˆıncˆat
det(aAB + bBA) = det(aBA + bAB). Demonstrat , i c˘a

det (x + a)AB + (b − x)BA = det (x + a)BA + (b − x)AB , ∀x ∈ C.
Daniel Jinga, Pites , ti