Exercitiu rezolvat 2 algebra clasa a 9 a

log a (6 x 2 + 5 x + 1) < 0

log a (6 x 2 + 5 x + 1) < log a 1


a) 6 x 2 + 5 x + 1 < 0
b) 6 x 2 + 5 x + 1 < 1



a) 6 x 2 + 5 x + 1 < 0

Δ = 25 – 4 * 6 * 1 (conform ecuatiei de gradul doi in care Δ = b 2 - 4ac)
Δ = 25 – 24
Δ = 1

x 1,2 = (-b +- √Δ )/2a

=> x 1 = -b + √Δ / 2a => -5 + 1 / 2*6 => -4/12 => -1/3
x 2 = -b - √Δ / 2a => -5 - 1 / 2*6 => -6/12 => -1/2


x
-∞
-1/2
-1/3
+∞
f(x)
++++++++++++
+++++0---------
---------0+++++
++++++++++++

x € (-∞,-1/2) U (-1/3, +∞) sau se mai poate scrie si x € (-∞,-0,50) U (-0,33, +∞)



b) 6 x 2 + 5 x + 1 < 1

6 x 2 + 5 x + 1 – 1 < 0
6 x 2 + 5 x < 0

6 x 2 + 5 x ≠ 0
6 x 2 + 5 x = 0
x(6x + 5) = 0

x 1 = 0
6x + 5 = 0 => 6x = -5 =>
x 2 = -5/6

x
-∞
-5 /6
0
+∞
f(x)
++++++++++++
+++++0---------
---------0+++++
++++++++++++
x € (-5/6, 0)


astfel =>

a) 6 x 2 + 5 x + 1 < 0         x € (-∞,-1 / 2) U (-1 / 3, +∞)


b) 6 x 2 + 5 x + 1 < 1          x € (-5/6, 0)



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