1) 5X^3*(3X - 5) \ 5X^3 * X^2 = ( 3X - 5) \ X^2
2) (3Y - 1)*(3Y + 1) \ 3*(3Y-1) = ( 3Y + 1) \ 3
3) ( A + 2)*(A + 2) \ ( 2 - A)*(2 + A) = ( A + 2) \ ( 2 - A )
{х-3у=2 и ху+у=6
{х=2+3у и ху+у=6
(2+3у)у+у=6
2у+3у²+у-6=0
3у²+3у-6=0
у²+у-2=0
D=1-4*(-2)=9
у₁=-1-3/2=-4/2=-2
у₂=-1+3/2=2/2=1
Если у₁=-2, то х₁=2+3*(-2)=2-6=-4
Если у₂=1, то х₂=2+3*1=5
Ответ: (-4;-2); (5;1)
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(3√а+3√в)²-(3√а-3√в)²=(3√а+3√в+3√а-3√в)(3√а+3√в-(3√а-3√в))=
=6√а(3√а+3√в-3√а+3√в)=6√а*6√в=36√(ав)