Question

решите систему неравенств
1-((1-х)/2)<4-((5+5х)/3), 2-((х+8)/4)>0

Answer 1

Ответ:

Объяснение:

$1-\frac{1-x}{2}<4-\frac{5+5x}{3} \\ 2-\frac{x+8}{4}>0$

6-3(1-x)<24-2(5x+5)

8-(x+8)>0

6-3+3x<24-10x-10

8-x-8>0

13x<11

x<0

Ответ: x∈(-∞;0)

Answer 2

Ответ:

$\left\{\begin{array}{l}1-\dfrac{1-x}{2}<4-\dfrac{5+5x}{3}\ \Big|\cdot 6\\2-\dfrac{x+8}{4}>0\ \Big|\cdot 4\end{array}\right\ \ \left\{\begin{array}{l}6-3(1-x)<24-2(5+5x)\\8-(x+8)>0\end{array}\right\\\\\\\left\{\begin{array}{l}6-3+3x<24-10-10x\\8-x-8>0\end{array}\right\ \ \left\{\begin{array}{l}13x<11\\-x>0\end{array}\right\ \ \left\{\begin{array}{l}x<\dfrac{11}{13}\\x<0\end{array}\right\ \ \Rightarrow \\\\\\\underline {\ x<0\ }\ \ \ ili\ \ \ \ \underline {\ x\in (-\infty ;0\, )\ }$