Question

розв'яжіть систему рівнянь ​

Answer 1

$\displaystyle\bf\\\left \{ {{ {x}^{2} + {y}^{2} = 25 } \atop {xy = - 12 }} \right. \\ \displaystyle\bf\\\left \{ {{x {}^{2} + ( - \frac{12}{x}) {}^{2} = 25 } \atop {y = - \frac{12}{x} }} \right. \\ \\ {x}^{2} + ( - \frac{12}{x} ) {}^{2} = 25 \\ {x}^{2} + \frac{144}{ {x}^{2} } - 25 = 0 \\ \frac{ {x}^{4} - 25 {x}^{2} + 144 }{ {x}^{2} } = 0 \\ {x}^{4} - 25 {x}^{2} + 144 = 0 \\ {x}^{2} = a \: , \: \: \: a \geqslant 0 \\ {a}^{2} - 25a + 144 = 0 \\ D = ( - 25) {}^{2} - 4 \times 144 = 625 - 576 = 49 \\ a _{1}= \frac{25 - 7}{2} = \frac{18}{2} = 9 \\ a_{2} = \frac{25 +7 }{2} = \frac{32}{2} = 16 \\ \\ {x}^{2} = 9 \\ {x}^{2} = 16 \\ \\ x_{1} = - 3 \\ x_{2} = 3 \\ x_{3} = - 4 \\ x_{4} = 4 \\ \\ y_{1} = - \frac{12}{ - 3} = 4 \\ y_{2} = - \frac{12}{3} = - 4 \\ y_{3} = - \frac{12}{ - 4} = 3 \\ y_{4} = - \frac{12}{4} = - 3$

Ответ: ( - 4 ; 3 ) , ( - 3 ; 4 ) , ( 3 ; - 4 ) , ( 4 ; - 3 )