Question

Решить систему уравнений
x+y=34,
log₂x+log₂y=6

Answer 1

$x+y=34 \\ log_{2}(xy)=6 \\ x=34-y \\ xy=64 \\ y(34-y)=64 \\ 34y- y^{2}-64=0 \\ y^{2}-34y+64=0 \\ y1=(34+30)/2=32 x1=2 \\ y2=(34-30)/2=2 x2=32$