Question

Алгебра. 30 баллов
Решите систему неравенств:
Фото в прикрепе

Answer 1

${3x}^{2} - x - 2 \leqslant 0 \\ 3x - 1 > 0 \\ \\ x∈\: [ - 1; \frac{2}{3} \: \:] \\ x > \frac{1}{3} \\ \\ x∈( \frac{1}{3} ; \frac{2}{3} ]$

Answer 2

Ответ:

$\left\{\begin{array}{l}3x^2+x-2\leq 0\\3x-1>0\end{array}\right\ \ \left\{\begin{array}{l}3\Big(x+1\Big)\Big(x-\dfrac{2}{3}\Big)\leq 0\\3x>1\end{array}\right\ \ \left\{\begin{array}{l}x\in [-1\ ;\ \frac{2}{3}\ ]\\x>\dfrac{1}{3}\end{array}\right\\\\\\\star \ \ 3x^2+x-2=0\ \ ,\ \ D=25\ \ ,\ \ x_1=-1\ ,\ x_2=\dfrac{2}{3}\ \ ,\\\\3x^2+x-2=3\Big(x+1\Big)\Big(x-\dfrac{2}{3}\Big)\ \ \star$

$\left\{\begin{array}{l}x\in \Big[-1\ ;\ \dfrac{2}{3}\ \Big]\\x\in \Big(\dfrac{1}{3}\ ;\, +\infty \, \Big)\end{array}\righ\ \ \ \Rightarrow \ \ \ \ \ \ x\in \Big(\, \dfrac{1}{3}\ ;\ \dfrac{2}{3}\ \Big]$