{∛x ·√y + ∛y ·√x =12 {∛(64/y) ·√y +∛y ·√(64/y) =12
{xy=64 (x≠0,y≠0) ⇒ x=64/y {x=64/y
{4·(√y/∛y)+8(∛y/√y)=12 пусть (√y/∛y)=t ⇒(∛y/√y)=1/t ⇒ 4t+8/t=12
4t²-12t+8=0
t²-3t+2=0 ⇔ t1=1
t2=2
t1=1 √y/∛y=1 √y=∛y ⇒y³=y², (y≠0) , y1=1 x1=64
проверка
y1=1 x1=64
{∛64 ·√1 + ∛1·√64 =12
{64·1=64 верно
t2=2 √y/∛y=2 √y=2∛y ⇒y³=64y², (y≠0) , y2=64 x2=1
проверка
y2=64 x2=1
{∛1 ·√64 + ∛64·√1 =12
{1·64=64 верно