Question

будь ласка допоможіть ​

Answer 1

Ответ:

$\displaystyle \emptyset$

Объяснение:

$\displaystyle \begin{cases}2-\frac{x-1}{3}\ge x\ \ \ |\cdot 3 \\ 7(x-3) > 5x-2\end{cases}\\\\\\\begin{cases}6-(x-1)\ge 3x\\ 7x-21 > 5x-2\end{cases}\\\\\\\begin{cases}6-x+1\ge 3x\\ 7x-5x > -2+21\end{cases}\\\\\\\begin{cases}-x-3x\ge-6-1\\ 2x > 19\ \ \ |:2\end{cases}\\\\\\\begin{cases}-4x\ge-7\ \ \ |:(-4)\\ x > 9,5\end{cases}\\\\\\\begin{cases}x\le 1,75 \\ x > 9,5\end{cases}\\\\\\x\in \emptyset$