Tìm x, y > 0 thỏa mãn x^3 + y^3 = 3xy - 1
Tìm x, y > 0 thỏa mãn x3 + y3 = 3xy – 1.
x3 + y3 = 3xy – 1
⇔ x3 + y3 – 3xy + 1 = 0
⇔ (x+y)3 – 3xy(x+y) – 3xy + 1 = 0
⇔ (x+y+1)(x2 + 2xy + y2 – x – y + 1 – 3xy) = 0
Suy ra:
\(\left[ \begin{array}{l}x + y + 1 = 0\\{x^2} + {\rm{ }}2xy{\rm{ }} + {\rm{ }}{y^2}--{\rm{ }}x{\rm{ }}--{\rm{ }}y{\rm{ }} + {\rm{ }}1{\rm{ }} - {\rm{ }}3xy = 0\end{array} \right.\)
Vì x, y > 0 nên x + y + 1 > 0
Xét x2 + 2xy + y2 – x – y + 1 – 3xy = 0
⇔ 2 (x2 + 2xy + y2 – x – y + 1 – 3xy) = 0
⇔ (x – y)2 + (x2 – 2x + 1) + (y2 – 2y + 1) = 0
⇔ (x – y)2 + (x – 1)2 + (y – 1)2 = 0
Suy ra:
\(\left\{ \begin{array}{l}x - y = 0\\x - 1 = 0\\y - 1 = 0\end{array} \right.\)hay x = y = 1.
Vậy x = y = 1.