Question

вот задача внизу

Answer 1

$left { {{x^2+y^2=48} atop {x+y=8=>x=8-y}} right\(8-y)^2+y^2=48\64-16y+y^2+y^2=48\2y^2-16y+16=0|:2\y^2-8y+8=0\y_{1,2}=frac{4^+_-sqrt{16-8}}{1}\y_1=4+2sqrt2 ;y_2=4-2sqrt2\x_1=4-2sqrt2 ;x_2=4+2sqrt2$
Пр-ка:
$(4-2sqrt2)^2+(4+2sqrt2)^2=48\16+8-16sqrt2+16+8+16sqrt2=48\48=48$