Question

Решите, пожалуйста. ​

Answer 1

Ответ:

$\bigg (12\dfrac{23}{30} \ ; \ +\infty \bigg )$

Пошаговое объяснение:

$\left|\begin{array}{ccc}15&2&x\\17&4&9\\18&6&4\end{array}\right|>1;$

$15 \cdot \left|\begin{array}{cc}4&9\\6&4\end{array}\right|-2 \cdot \left|\begin{array}{cc}17&9\\18&4\end{array}\right|+x \cdot \left|\begin{array}{cc}17&4\\18&6\end{array}\right|>1;$

$15 \cdot (4 \cdot 4-9 \cdot 6)-2 \cdot (17 \cdot 4-9 \cdot 18)+x \cdot (17 \cdot 6-4 \cdot 18)>1;$

$15 \cdot (16-54)-2 \cdot (68-162)+x \cdot (102-72)>1;$

$15 \cdot (-38)-2 \cdot (-94)+x \cdot 30>1;$

$-570+188+30x>1;$

$30x-382>1;$

$30x>382+1;$

$30x>383;$

$x>\dfrac{383}{30} \ ;$

$x>12\dfrac{23}{30} \ ;$

$x \in \bigg (12\dfrac{23}{30} \ ; \ +\infty \bigg );$