Question

Решение систем уравнений второй степени

Answer 1

$begin{cases}x=y+26,\ xy=3800;end{cases}, begin{cases}x=y+26,\ (y+26)y=3800;end{cases},begin{cases}x=y+26,\ y^2+26y-3800=0;end{cases}\ begin{cases}x=y+26,\ (y+76)(y-50)=0;end{cases}, begin{cases}x_1=-50,,x_2=76,\ y_1=-76,,y_2=50.end{cases}$
Ответ: 76

$begin{cases}x+y=39(x-y),\ x^2-y^2=39;end{cases}, begin{cases}38x=40y,\ (x-y)(x+y)=39;end{cases}\ begin{cases}y=0,95x,\ (x-0,95x)(x+0,95x)=39;end{cases},begin{cases}y=0,95x,\ 0,05xcdot 1,95x=39;end{cases}\ begin{cases}y=0,95x,\ 0,0975x^2=39;end{cases},begin{cases}y=0,95x,\ x^2=400;end{cases},begin{cases}y_{1,2}=pm 19,\ x_{1,2}=pm 20.end{cases}$
Ответ:19