8 - 9 классы. Срочно! №95. Хотя бы один пример. Пожалуйста!!! Фотография прикреплена.

Question 1

Question 2

Решите задачу:

$\left \{ {{2x^2-3xy-19y^2=25\, |\cdot (-10)} \atop {x^2-6y^2=250}} \right. \oplus \left \{ {{2x^2-3xy-19y^2=25y=2} \atop {-19x^2+30xy+184y^2=0}} \right. \\\\19x^2-30xy-184y^2=0\; |:y^2\; (y\ne 0)\\\\19\cdot ( \frac{x}{y})^2 -30\cdot \frac{x}{y}-184=0\\\\t= \frac{x}{y} \; ,\; \; 19t^2-30t-184=0\\\\D/4=15^2+19\cdot 184=3721=61^2\\\\t_1= \frac{15-61}{19}=-\frac{46}{19}\; ,\; \; t_2= \frac{15+61}{19}=\frac{76}{19} =4\\\\a)\; \; \frac{x}{y}=-\frac{46}{19} \; ,\; \; x=- \frac{46}{19}\cdot y$

$x^2-6y^2=(- \frac{46y}{19})^2-6y^2= \frac{2116y^2}{361}-6y^2=\frac{2116-114}{361}\cdot y^2= \frac{2002}{361}\cdot y^2\\\\ \frac{2002}{361}\cdot y^2=250\; \; \to \; \; y^2= \frac{250\cdot 361}{2002}= \frac{45125}{1001}\; ,\\\\y_{1,2}=\pm \frac{95\sqrt5}{\sqrt{1001}}=\pm 95\sqrt{\frac{5}{1001}}$

$x=-\frac{46}{19} \cdot y=- \frac{46}{19}\cdot \Big (\pm 95\sqrt{\frac{5}{1001}} \Big )=\mp \frac{4370}{19}\cdot \sqrt{\frac{5}{1001}}$

$b)\; \; \frac{x}{y} =4\; ,\; \; x=4y\\\\x^2-6y^2=(4y)^2-6y^2=16y^2-6y^2=10y^2\\\\10y^2=250\; \; \to \; \; y^2=25\; ,\; \; y_{3,4}=\pm 5\\\\x=4\cdot y=4\cdot (\pm 5)=\pm 20$

$Otvet:\; \; \Big (\mp \frac{4370}{19}\sqrt{\frac{5}{1001}}\; ;\; \pm 95\sqrt{\frac{5}{1001}}\Big )\; ,\Big (\pm 5;\pm 20\Big )\; .$