ОДЗ
{x+y>0⇒x>-y
{x-y>0⇒x>y
{log(2)(x+y)+[2log(2)(x-y)]/2=5⇒log(2)(x+y)+log(2)(x-y)=5⇒log(2)(x²-y²)=5
{3*3^log(3)(x-y)²=48⇒3^(log(3)(x-y)²=16⇒(x-y)²=16
1)x-y=-4
x=y-4
log(2)(y²-8y+16-y²)=5
log(2)(-8y+16)=5
-8y+16=32
-8y=16
y=-2
x=-2-4
x=-6
x2)x-y=4
x=y+4
log(2)(y²+8y+16-y²)=5
log(2)98y+16)=5
8y+16=32
8y=16
y=2
x=2+4
x=6
Ответ (6;2)