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)
Exercitii rezolvate clasa a 9 a
Probleme si exercitii rezolvate
Exercitii rezolvate clasa a 9 a
Probleme si exercitii rezolvate