Помогите пожалуйста решить систему уравнений!!! Очень срочно надо.

Question 1

Question 2

Какую из четырёх систем решить надо ?

Question 3

две последние

Question 4

Решите задачу:

$1)\; \; \left \{ {{ \frac{7}{\sqrt{x-2}} - \frac{4}{\sqrt{y+2}} = \frac{5}{3} } \atop { \frac{5}{\sqrt{x-2}} + \frac{3}{\sqrt{y+2}} = \frac{13}{6} }} \right. \; \; Zamena:\; \; u= \frac{1}{\sqrt{x-2}}\; ;\; \; v= \frac{1}{\sqrt{y+2}} \\\\ \left \{ {{7u-4v=\frac{5}{3}} \atop {5u+3v=\frac{13}{6}}} \right. \; \left \{ {{21u-12v=5\; |\cdot 3} \atop {30u+18v=13\; |\cdot 2}} \right. \; \oplus \; \left \{ {{21u-12v=5} \atop {123u=41}} \right. \; \left \{ {{12v=21u-5} \atop {u=\frac{1}{3}}} \right.$

$12v=21\cdot \frac{1}{3}-5=7-5=2\; \; \to \; \; \; v=\frac{2}{12}=\frac{1}{6}\\\\u=\frac{1}{\sqrt{x-2}}=\frac{1}{3}\quad \to \quad \sqrt{x-2}}=3\; ,\; \; x-2=9\; ,\; \; x=11\\\\v=\frac{1}{\sqrt{y+2}}=\frac{1}{6}\quad \to \quad \sqrt{y+2}=6\; ,\; y+2=36\; ,\; \; y=34\\\\Otvet:\; \; (11;34)\; .$

$2)\; \; \left \{ {{\sqrt{2x-3y}+\sqrt{2x+3y}=10} \atop {\sqrt{4x^2-9y^2}=16}} \right. \; \left \{ {{\sqrt{2x-3y}+\sqrt{2x+3y}=10=2} \atop {\sqrt{2x-3y}\, \cdot \, \sqrt{2x+3y}=16}} \right. \\\\Zamena:\; \; u=\sqrt{2x-3y}\; ;\; \; v=\sqrt{2x+3y}\\\\ \left \{ {{u+v=10} \atop {u\cdot v=16}} \right. \; \left \{ {{u=10-v} \atop {(10-v)\cdot v=16}} \right. \; \left \{ {{u=10-v} \atop {v^2-10v+16=0}} \right. \; \left \{ {{u_1=8\; ,\; u_2=2} \atop {v_1=2\; ,\; v_2=8}} \right.$

$a)\; \; \left \{ {{\sqrt{2x-3y}=8} \atop {\sqrt{2x+3y}=2}} \right. \; \left \{ {{2x-3y=64} \atop {2x+3y=4}} \right. \; \left \{ {{4x=68} \atop {6y=-60}} \right. \; \left \{ {{x=17} \atop {y=-10}} \right. \\\\b)\; \; \left \{ {{\sqrt{2x-3y}=2} \atop {\sqrt{2x+3y}=8}} \right. \; \left \{ {{2x-3y=4} \atop {2x+3y=64}} \right. \; \left \{ {{4x=68} \atop {6y=60}} \right. \; \left \{ {{x=17} \atop {y=10}} \right. \\\\Otvet:\; \; (17;-10)\; ,\; (17;10)\; .$