Question

помогите режить пожалуйста​

Answer 1

$2)\; \; \left \{ {{x^2+y^2=5\qquad } \atop {x^2+y^2+3xy=-1|\cdot 5}} \right.\; \oplus \left \{ {{x^2+y^2=5\qquad } \atop {6x^2+6y^2+15xy=0}} \right.\\\\6x^2+6y^2+15xy=0\, |:y^2\ne 0\\\\6(\frac{x}{y})^2+15\frac{x}{y}+6=0\; \; ,\; \; t=\frac{x}{y}\; \; \to \; \; 6t^2+15t+4=0\; ,\\\\2t^2+5t+2=0\; \; ,\; \; D=25-16=9\; ,\; \; t_1=-2\; ,\; \; t_2=-\frac{1}{2}\\\\a)\; \; \frac{x}{y}=-2\; \; ,\; \; x=-2y\; \; \to \; \; x^2+y^2=(-2y)^2+y^2=5y^2=5\; ,\\\\y^2=1\; \; ,\; \; y=\pm 1\; \; \to \; \; x=\mp 2$

$b)\; \; \frac{x}{y}=-\frac{1}{2}\; \; ,\; \; y=-2x\; \; ,\; \; x^2+y^2=5y^2=3\; ,\\\\y^2=1\; \; ,\; \; y=\pm 1\; \; \to \; \; x=\mp 2\\\\Otvet:\; \; (-2,1)\; ,\; (2,-1)\; ,\; (1,-2)\; ,\; (-1,2)\; .$

$3)\; \; \left \{ {{x+y=3} \atop {x^3+y^3=9}} \right.\; \left \{ {{x+y=9\qquad \qquad \; \; } \atop {(x+y)(x^2-xy+y^2)=9}} \right.\; \left \{ {{x+y=3\qquad \quad } \atop {3(x^2-xy+y^2)=9}} \right.\; \left \{ {{y=3-x\qquad } \atop {x^2-xy+y^2=3}} \right.\\\\\left \{ {{y=3-x\qquad \qquad \; \; \; } \atop {x^2-x(3-x)+(3-x)^2=3}} \right. \\\\x^2-3x+x^2+9-6x+x^2=3\\\\3x^2-9x+6=0\; \; ,\; \; x^2-3x+2=0\; \; ,\; \; x_1=1\; ,\; x_2=2\; \; (teor.\; Vieta)\\\\y_1=3-1=2\; \; ,\; \; y_2=3-2=1\\\\Otvet:\; \; (1,2)\; ,\; (2,1)\; .$

$5)\; \; \left \{ {{x^2-y^2=15} \atop {x-y=3}} \right.\; \left \{ {{(x-y)(x+y)=15} \atop {x-y=3}} \right.\; \left \{ {{3(x+y)=15} \atop {x-y=3}} \right.\; \left \{ {{x+y=5} \atop {x-y=3}} \right.\; \oplus \; \left \{ {{2x=8} \atop {y=x-3}} \right.\\\\\left \{ {{x=4} \atop {y=1}} \right.\\\\Otvet:\; \; (4,1)\; .$

$4)\; \; \left \{ {{x^3+3xy^2=36\, |\cdot 28} \atop {y^3+3x^2=28|\cdot (-36)}} \right.\; \oplus \; \left \{ {{7x^3+3xy^2=36\qquad \qquad \quad } \atop {28x^3+84xy^2-36y^3-108x^2y=0}} \right.\\\\28x^3-108x^2y+84xy^2-36y^3=0\, |:y^3\ne 0\\\\t=\frac{x}{y}\; ,\; \; \; 28t^3-108t^2+84t-36=0\\\\t=3\; -\; koren\; (podbor)\; \; \to \\\\28t^3-108t^2+84t-30=(t-3)(28t^2-24t+12)\\\\28t^2-24t+12=0\; \; \to \\\\7t^2-6t+3=0\; ,\; \; D=-48<0\; \; \to \; \; t\in \varnothing \\\\t=\frac{x}{y}=3\; \; ,\; \; x=3y$

$x^3+3xy^2=27y^3+9y^3=36y^3\\\\36y^3=36\\\\y^3=1\\\\y=1\\\\x=3y=3\\\\Otvet:\; \; (3,1)\; .$

Answer 2

Ответ: во вложении Объяснение: