Giải hệ phương trình: $\left\{\begin{array}{l}{x}^2+y^2=65\\ \left(x-1\right)\left(y-1\right)=18\end{array}\right.$ .

Ta có: (x – 1)(y – 1) = 18

⇔ xy – (x + y) + 1 = 18

⇔ xy – (x + y) = 17

⇔ xy = x + y + 17

Lại có: x² + y² = 65

⇔ (x + y)² – 2xy = 65

⇔ (x + y)² – 2(x + y + 17) – 65 = 0

⇔ (x + y)² – 2(x + y) – 99 = 0

⇔ $\left[\begin{array}{l}x+y=11\\ x+y=-9\end{array}\right.$

Nếu x + y = 11 thì xy = 28. Suy ra:

x(11 – x) – 28 = 0

⇔ – x² + 11x – 28 = 0

⇔ $\left[\begin{array}{l}x=4\\ x=7\end{array}\right.$  ⇒ $\left[\begin{array}{l}y=7\\ y=4\end{array}\right.$

Nếu x + y = -9 thì xy = 8. Suy ra:

x(-9 – x) = 8

⇔ x² + 9x + 8 = 0

⇔ $\left[\begin{array}{l}x=-1\\ x=-8\end{array}\right.$  ⇒ $\left[\begin{array}{l}y=-8\\ y=-1\end{array}\right.$ .