Question

[Умоляю!]
Решите систему неравенств на фото!
Даю (100) баллов!

Answer 1

Пусть x = 3k, y = 3m. Тогда

$\left \{ {{2*(3k)^2+4*3k*3m+11*(3m)^2\leq 1} \atop {4*3k+7*3m\geq 3}} \right. \\\left \{ {{18k^2+36km+99m^2\leq 1} \atop {4k+7m\geq 1}} \right. \\\left \{ {{18k^2+36km+99m^2\leq 1} \atop {16k^2+56km+49m^2\geq 1}} \right. \\\left \{ {{18k^2+36km+99m^2\leq 1} \atop {-16k^2-56km-49m^2\leq -1}} \right. \Rightarrow 18k^2-16k^2+36km-56km+99m^2-49m^2\leq 1-1\\2k^2-20km+50m^2\leq 0\\k^2-10km+25m^2\leq 0\\(k-5m)^2\leq 0 \Leftrightarrow k-5m=0\\k=5m$

$\left \{ {{18*(5m)^2+36*5m*m+99m^2\leq 1} \atop {4*5m+7m\geq 1}} \right. \\\left \{ {{450m^2+180m^2+99m^2\leq 1} \atop {20m+7m\geq 1}} \right. \\\left \{ {{729m^2\leq 1} \atop {27m\geq 1}} \right. \left \{ {{m^2\leq\frac{1}{729} } \atop {m\geq \frac{1}{27}}} \\\right. \\\left \{ {{-\frac{1}{27}\leq m\leq\frac{1}{27}} \atop {m\geq\frac{1}{27}}} \right.\Rightarrow m=\frac{1}{27},\ k=5m=\frac{5}{27}\\x=3k=\frac{5}{9},\ y=3m=\frac{1}{9}$

Ответ: $(\frac{5}{9};\frac{1}{9})$