Question

методом додавання
завчасно дякую​

Answer 1

$x {}^{2} + y {}^{2} = 25 \\ y = - \frac{12}{x}$

$x {}^{2} + ( - \frac{12}{x} ) {}^{2} = 25 \\ x {}^{2} + \frac{144}{x {}^{2} } = 25 \\ x {}^{2} + \frac{144}{x {}^{2} } - 25 = 0 \\ \frac{x {}^{4} + 144 - 25x {}^{2} }{x {}^{2} } = 0 \\ x {}^{4} + 144 - 25x {}^{2} = 0 \\ x {}^{4} - 25x {}^{2} + 144 = 0$

$t = x {}^{2}$

$t {}^{2} - 25t + 144 = 0$

$t1 = 9 \\ t2 = 16$

сделать обратную замену:

$x {}^{2} = 9 \\ x {}^{2} = 16$

$x = 3 \\ x = - 3$

$x = 4 \\ x = - 4$

подставим данные значения в xy = -12

$- 3y = - 12 \\ 3y = - 12 \\ - 4y = - 12 \\ 4y = 12$

$y = 4 \\ y = - 4 \\ y = 3 \\ y = - 3$

$(x1.y1) = ( - 3.4) \\ (x2.y2) = (3. - 4) \\ (x3.y3) = ( - 4.3) \\ (x4.y4) = (4. - 3)$