Question

АЛГЕБРА! Только 1 и 3. СРОЧНО!
ЗАДАНИЕ : Знайдіть множину розв'язків системи нерівностей ​

Answer 1

Ответ:

$\displaystyle 1)\ (-\frac{1}{2};-\frac{3}{8})$

$\displaystyle 3)\ \emptyset$

Объяснение:

$\displaystyle 1)\\\\\begin{cases}\frac{2x-3}{5}-\frac{4x-9}{6} > 1\ \ \ |\cdot 30 \\5(x-1)+7(x+2) > 3\end{cases}\\\\\\\begin{cases}6(2x-3)-5(4x-9) > 30\\5x-5+7x+14 > 3\end{cases}\\\\\\\begin{cases}12x-18-20x+45 > 30\\5x+7x > 3+5-14\end{cases}\\\\\\\begin{cases}12x-20x > 30+18-45\\12x > -6\ \ \ |:12\end{cases}\\\\\\\begin{cases}-8x > 3\ \ \ |:(-8)\\x > -\frac{1}{2}\end{cases}\\\\\\\begin{cases}x < -\frac{3}{8}\\x > -\frac{1}{2}\end{cases}\\\\\\x\in(-\frac{1}{2};-\frac{3}{8})$

$\displaystyle 3)\\\\\begin{cases}(x-6)^2 < (x-2)^2-8 \\ 3(2x-1)-8 < 34-3(5x-9)\end{cases}\\\\\\\begin{cases}x^2-12x+36 < x^2-4x+4-8 \\ 6x-3-8 < 34-15x+27\end{cases}\\\\\\\begin{cases}x^2-12x-x^2+4x < 4-8 -36\\ 6x+15x < 34+27+3+8\end{cases}\\\\\\\begin{cases}-8x < -40\ \ \ |:(-5)\\ 21x < 72\ \ \ |:21\end{cases}\\\\\\\begin{cases}x > 5\\ x < \frac{24}{7}\end{cases}\\\\\\\begin{cases}x > 5\\ x < 3\frac{23}{7}\end{cases}\\\\\\x\in\emptyset$