Решите систему уравнений:

Решите задачу:

$\left \{ {{x-y=1} \atop {xy=240}} \right. \\ \left \{ {{x = 1 + y} \atop {(1+y) * y=240}} \right. \\ (1 + y) * y = 240 \\ y + y^2 - 240 = 0 \\ y^2 + y - 240 = 0 \\ D = 1^2 - 4 * (-240) = 1+960 = 961 \\ y_1_,_2 = \frac{-1 б \sqrt{D} }{2} = \frac{-1 б 31 }{2} \\ y_1 = 15 \\ y_2 = -16 \\ x = 1+y \\ x_1 = 1+y_1 = 1+15 = 16 \\ x_2 = 1+y_2 = 1+ (-16) = 1 - 16 = -15 \\ (16;15) (-15; -16)$

$\left \{ {{x^2+y^2=65} \atop {2x- y=15}} \right \\ \left \{ {{x^2 +y^2=65} \atop {y = 2x - 15}} \right. . \\ x^2 + (2x - 15)^2 = 65 \\ x^2 + 4x^2 - 60x + 225 - 65 = 0 \\ 5x^2 - 60 x + 160 = 0 \\ D = (-60)^2 - 4 * 5 * 160 = 3600 - 3200 = 400 \\ x_1,_2 = \frac{60 б \sqrt{D} }{5*2} = \frac{60б 20}{10} \\ x_1 = 8 \\ x_2 = 4 \\ y = 2x - 15 \\ y_1 = 2x_1 - 15 = 2*8 - 15 = 1 \\ y_2 = 2x_2 - 15 = 2 * 4 - 15 = -7 \\ (8;1)(4;-7)$