Алгебра: Решите системы уравнений {x+y=1 x^2+y^2=25 {x+y=3 x^2+y^2=29

Решите системы уравнений {x+y=1 x^2+y^2=25 {x+y=3 x^2+y^2=29

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Ответ:

Алгебра111111111

03.10.2020 19:12

$1)\quad \left \{ {{x+y=1} \atop {x^2+y^2=25}} \right. \; \left \{ {{(x+y)^2=1} \atop {x^2+y^2=25}} \right. \; \left \{ {{x^2+y^2+2xy=1} \atop {x^2+y^2=25}} \right. \; \left \{ {{25+2xy=1} \atop {x^2+y^2=25}} \right. \\\\ \left \{ {{2xy=-24} \atop {x^2+y^2=25}} \right. \; \left \{ {{xy=-12} \atop {x^2+y^2=25}} \right. \; \left \{ {{y=\frac{-12}{x}} \atop {x^2+\frac{144}{x^2}=25}} \right. \; \left \{ {{y=-\frac{12}{x}} \atop {x^4-25x^2+144=0\; ,\; x\ne 0}} \right.$

$\left \{ {{y=-\frac{12}{x}} \atop {x^2=9\; ,\; x^2=16}} \right. \; \left \{ {{y_{1,2}=\mp 4\; ,\; x_{3,4}=\mp 3} \atop {x_{1,2}=\pm 3\; ,\; x_{3,4}=\pm 4}} \right. \\\\Otvet:\; (3,-4)\; ,\; (-3,4)\; ,\; (4,-3)\; ,\; (-4,3)\; .$

$2)\quad \left \{ {{x+y=3} \atop {x^2+y^2=29}} \right. \; \left \{ {{(x+y)^2=9} \atop {x^2+y^2=29}} \right. \; \left \{ {{9=29+2xy} \atop {x^2+y^2=29}} \right. \; \left \{ {{xy=-10} \atop {x^2+y^2=29}} \right. \\\\ \left \{ {{y=-\frac{10}{x}} \atop {x^2+\frac{100}{x^2}=29}} \right. \; \left \{ {{y=-\frac{10}{x}} \atop {x^4-29x^2+100=0,\; x\ne 0}} \right. \; \left \{ {{y=-\frac{10}{x}} \atop {x^2=4\; ,\; x^2=25}} \right.$

$\left \{ {{y=-\frac{10}{x}} \atop {x_{1,2}=\pm 2\; ,\; x_{3,4}=\pm 5} \right. \\\\$ $\left \{ {{y_{1,2}=\mp 5\; ,\; y_{3,4}=\mp 2} \atop {x_{1,2}=\pm 2\; ,\; x_{3,4}=\pm 5}} \right.$

$Otvet:\; \; (2,-5)\; ,\; (-2,5)\; ,\; (5,-2)\; ,\; (-5,2)\; .$