Question

Rezolvati in R sistemul de inecuatii:

a) {(9x+3)(x-4) > 9x^2 + x + 5,

{2x-3-(x-3) mai mic sau egal ca 5


b) {12x^2 - (x-3)(6x + 1) > x

{(5x-1)(5x+1) - 25x^2 > x-6

Answer 1

(9x + 3)(x - 4) > 9x² + x + 5

9x² - 33x - 12 > 9x² + x + 5

9x² - 9x² - 33x - x > 12 + 5

-34x > 17

34x < - 17

x < - 17/34

x < - 1/2

x ∈ (-oo ; - 1/2)

2x - 3 - (x - 3) ≤ 5

2x - 3 - x + 3 ≤ 5

x ≤ 5

x ∈ (- oo ; 5]

12x² - (x - 3)(6x + 1) > x

12x² - (6x² - 17x - 3) > x

12x² - 6x² + 17x + 3 > x

6x² + 17x - x + 3 > 0

6x² + 16x + 3 > 0

rezolvam ecuatia

6x² + 16x + 3 = 0

x = (- 16±√256-72)/12

= (- 16±√184)/12

= (- 16±2√46)/12

= (-8±√46)/6

avem

(x - (-8 + √46)/6)(x - (-8+√46)/6) > 0

x ∈ (-oo ; (-8+√46)/6) ∪ ((-8 + √46)/6) ; +oo)

(5x - 1)(5x + 1) - 25x² > x - 6

5x² - 1 - 25x² + 6 > x

x < 5

x ∈ (-oo ; 5)

Answer 2

$\it b)\ \begin{cases}\it 12x^2-(2x-3)(6x+1)>x\\ \\ \it (5x-1)(5x+1)-25x^2>x-6\end{cases}\Leftrightarrow \begin{cases}\it 12x^2--12x^2-2x+18x+3>x\\ \\ \it 25x^2-1-25x^2>x-6\end{cases}\Leftrightarrow\\ \\ \\ \Leftrightarrow \begin{cases}\it 15x>-3\\ \\ \it x<5\end{cases}\Leftrightarrow \begin{cases}\it x>-\dfrac{1}{5}\\ \\ \it x<5\end{cases}\Leftrightarrow\ \ S=\Big(-\dfrac{1}{5},\ 5\Big)$