341 (1), 343 (1) Заранее благодарю)

Question 1

Question 2

Решите задачу:

$1)\; \; \left \{ {{2\sqrt{3y+x}-\sqrt{6y-x}\; =x} \atop {\sqrt{3y+x}+\sqrt{6y-x}\; =3y}} \right. \; \oplus \; \left \{ {{3\sqrt{3y+x}=3y+x} \atop {\sqrt{3y+x}+\sqrt{6y-x}=3y}} \right. \; \left \{ {{3\sqrt{3y+x}-(3y+x)=0} \atop {\sqrt{6y-x}=3y-\sqrt{3y+x}}} \right. \\\\a)\; \; 3\sqrt{3y+x}-(3y+x)=0\\\\\sqrt{3y+x}\cdot (3-\sqrt{3y+x})=0\; \; \Rightarrow \\\\\sqrt{3y+x}=0\; \; ili\; \; \; \sqrt{3y+x}=3\\\\3y+x=0\; \; \; ili\; \; \; 3y+x=9\\\\x=-3y\; \; \; ili\; \; \; x=9-3y$

$b)\; \; \left \{ {{x=-3y} \atop {\sqrt{6y-3}=3y-0}} \right. \; \left \{ {{x=-3y} \atop {6y-3=9y^2\; ,\; y\ \textgreater \ 0}} \right. \; \left \{ {{x=-3y} \atop {9y^2-6y+3=0}} \right. \\\\3y^2-2y+1=0\; ,\\\\D/4=1-3=-2\ \textless \ 0\; \; \to \; \; 3y^2-2y+1\ne 0\; \; (3y^2-2y+1\ \textgreater \ 0)\\\\net\; reshenij\\\\c)\; \; \left \{ {{x=9-3y} \atop {\sqrt{6y-x}=3y-3}} \right. \; \left \{ {{x=9-3y} \atop {6y-(9-3y)=(3y-3)^2}} \right. \; \left \{ {{x=9-3y} \atop {9y-9=9y^2-18y+9}} \right. \\\\ \left \{ {{x=9-3y} \atop {9y^2-27y+18=0}} \right. \; \left \{ {{x=9-3y} \atop {y^2-3y+2=0}} \right. \; \left \{ {{x_1=6\; ,\; x_2=3} \atop {y_1=1\; ,\; y_2=2}} \right. \\\\(6,1)\; ,\; \; (3,2)$

$Proverka:\\\\(6,1):\; \; \left \{ {{2\sqrt{9}-\sqrt{0}=6} \atop {\sqrt{9}+\sqrt{0}=3}} \right. \\\\(3,2):\; \; \left \{ {{2\sqrt{9}-\sqrt{9}=3} \atop {\sqrt{9}+\sqrt{9}=6}} \right.\\\\Otvet:\; \; (6,1)\; ,\; (3,2).$

$2)\; \; \left \{ {{x^2+y^2=5} \atop {x+y=2}} \right. \; \left \{ {{x^2+y^2=5} \atop {(x+y)^2=4}} \right. \; \left \{ {{x^2+y^2=5} \atop {x^2+y^2+2xy=4}} \right. \; \left \{ {{x^2y^2=5} \atop {5+2xy=4}} \right. \\\\ \left \{ {{x^2+y^2=5} \atop {xy=-\frac{1}{2}}} \right. \; \left \{ {{x^2+\frac{1}{4x^2}=5} \atop {y=-\frac{1}{2x}}} \right. \; \left \{ {{4x^4-20x^2+1=0} \atop {y=-\frac{1}{2x}}} \right. \\\\4x^4-20x^2+1=0\\\\t=x^2 \geq 0\; ,\; \; \; 4t^2-20t+1=0\; ,\; \; D/4=100-4=96$

$t_1=\frac{10-4\sqrt6}4}= \frac{5-2\sqrt6}{2}=2,5-\sqrt6 \; ,\; \; t_2= \frac{5+2\sqrt6}{2} = 2,5+\sqrt6 \\\\ x^2=2,5+\sqrt6\; \; \to \; \; \; x=\pm \sqrt{2,5+\sqrt6}=\pm \frac{\sqrt{5+2\sqrt6}}{\sqrt2 }\\\\y=-\frac{1}{\pm 2\sqrt{2,5+\sqrt6}}\; \; \to \\\\y_1=-\frac{\sqrt2}{-2\sqrt{5+2\sqrt6}}= \frac{1}{\sqrt{2(5+2\sqrt6)}} = \frac{1}{\sqrt{10+4\sqrt6}} = \frac{\sqrt{10-4\sqrt6}}{\sqrt{100-96}}= \frac{\sqrt{10-4\sqrt6}}{2} \\\\y_2=-\frac{\sqrt2}{2\sqrt{5+2\sqrt6}}=- \frac{\sqrt{10-4\sqrt6}}{2}$

$x^2=\frac{5-2\sqrt6}{2}\; \; \to \; \; x_{3,4}=\pm \frac{\sqrt{5-2\sqrt6}}{\sqrt2} \; \; \to \\\\y_{3,4}=\mp \frac{1}{\sqrt2\cdot \sqrt{5-2\sqrt6}}=\mp \frac{\sqrt{10+4\sqrt6}}{2}$

$Otvet:\; \; \Big ( -\frac{\sqrt{5+2\sqrt6}}{\sqrt2} \; ;\; \frac{\sqrt{10-4\sqrt6}}{2} \Big )\; ,\; \Big ( \frac{\sqrt{5+2\sqrt6}}{\sqrt2} \; ;\; -\frac{\sqrt{10-4\sqrt6}}{2} \Big )\; ,$

$\Big ( \frac{\sqrt{5-2\sqrt6}}{\sqrt2};-\frac{\sqrt{10+4\sqrt6}}{2}\Big )\; ,\; \Big (-\frac{\sqrt{5-2\sqrt6}}{\sqrt2}\; ;\; \frac{\sqrt{10+4\sqrt6}}{2}\Big )\; .$