Question

2,3. Решите пожалуйста. Срочно

Answer 1

Объяснение:

2)  Симметрическая система .

$\left \{ {{x^2+3xy+y^2=-1} \atop {x^3+y^3=7}} \right.\; \; \quad zamena:\; u=x+y\; ,\; \; v=xy\; ,\\\\\\u^2=(x+y)^2=x^2+y^2+2xy=x^2+y^2+3v\; \; \to \; \; x^2+y^2=u^2-2v\\\\u^3=(x+y)^3=x^3+y^3+3xy(x+y)=x^3+y^3+3uv\; \to \; x^3+y^3=u^3-3uv\\\\\\\left \{ {{u^2-2y+3v=-1} \atop {u^3-3uv=7}} \right.\; \; \left \{ {{u^2+v=-1} \atop {u^3-3uv=7}} \right.\; \; \left \{ {{v=-u^2-1} \atop {u^3-3u(-u^2-1)=7}} \right.\; \; \left \{ {{v=-u^2-1} \atop {4u^3+3u-7=0}} \right.\\\\4u^3+3u-7=0\; \; \to \; \; podborom:\; \; u=1$

$4u^3+3u-7=(u-1)(4u^2+4u+7)=0\\\\u-1=0\; \; \to \; \; u=1\\\\4u^2+4u+7=0\; \; \to \; \; D=16-16\cdot 7<0\; \; \to \; \; net\; reshenij\; \; dejstvitelnux\\\\v=-u^2-1=-1^2-1=-2\\\\\\\left \{ {{x+y=1} \atop {xy=-2}} \right.\; \; \left \{ {{y=1-x} \atop {x(1-x)=-2}} \right.\; \; \left \{ {{y=1-x} \atop {x^2-x-2=0}} \right.\; \; \left \{ {{y_1=2\; ,\; \; y_2=-1} \atop {x_1=-1\; ,\; \; x_2=2}} \right.\\\\Otvet:\; \; (-1,2)\; ,\; \; (2,-1)\; .$

3)  Однородная система.

$\left \{ {{x^2-2xy+y^2=25\; |\cdot (-11)} \atop {2x^2-2xy-y^2=11\; |\cdot 25}} \right.\; \oplus \; \left \{ {{x^2-2xy+y^2=25\; \; \; \; \; } \atop {14x^2-28xy-36y^2=0}} \right.\\\\\\14x^2-28xy-36=0\; \; \to \; \; \; 7x^2-14xy-18y^2=0\, |:y^2\ne 0\\\\\\7\cdot (\frac{x}{y})^2-14\cdot \frac{x}{y}-18=0\; \; ,\; \; \; t=\frac{x}{y}\; \; \to \; \; 7t^2-14t-18=0\; ,\\\\D/4=7^2+7\cdot 18=175=25\cdot 7\; \; ,\; \; \; t_1=\frac{7-5\sqrt7}{7}=1-\frac{5}{\sqrt7}\; ,\; t_2=1+\frac{5}{\sqrt7}\; ,$

$a)\; \; \frac{x}{y}=1-\frac{5}{\sqrt7}\; \; \to \; \; x=(1-\frac{5}{\sqrt7})\cdot y\; ,\\\\x^2-2xy+y^2=(1-\frac{10}{\sqrt7}+\frac{25}{7})\cdot y^2-2(1-\frac{5}{\sqrt7})\cdot y^2+y^2=\\\\=(1-\frac{10}{\sqrt7}+\frac{25}{7}-2+\frac{10}{\sqrt7}+1)\cdot y^2=\frac{25}{7}\cdot y^2\; \; ,\; \; \; \frac{25}{7}\cdot y^2=25\; ,\\\\y^2=7\; \; \Rightarrow \; \; \; y_{1,2}=\pm \sqrt7\\\\x_1=(1-\frac{5}{\sqrt7})\cdot (-\sqrt7)=5-\sqrt7\; ,\; x_2=(1-\frac{5}{\sqrt7})\cdot \sqrt7=-5+\sqrt7$

$b)\; \; \frac{x}{y}=1+\frac{5}{\sqrt7}\; \; \to \; \; x=(1+\frac{5}{\sqrt7})\cdot y\; ,\\\\x^2-2xy+y^2=(1+\frac{10}{\sqrt7}+\frac{25}{7})\cdot y^2-2(1+\frac{5}{\sqrt7})\cdot y^2+y^2=\\\\=(1+\frac{10}{\sqrt7}+\frac{25}{7}-2-\frac{10}{\sqrt7}+1)\cdot y^2=\frac{25}{7}\cdot y^2\; \; ,\; \; \; \frac{25}{7}\cdot y^2=25\; ,\\\\y^2=7\; \; \Rightarrow \; \; \; y_{3,4}=\pm \sqrt7\\\\x_3=(1+\frac{5}{\sqrt7})\cdot (-\sqrt7)=-5-\sqrt7\; ,\; x_2=(1+\frac{5}{\sqrt7})\cdot \sqrt7=5+\sqrt7\\\\\\Otvet:\; \; (5-\sqrt7\, ;\, -\sqrt7)\; ,\; (-5+\sqrt7\, ;\, \sqrt7)\; ,\; (-5-\sqrt7\, ;\, -\sqrt7)\; ,\; (5+\sqrt7\, ;\, \sqrt7)\; .$