\(^{x^2-xy+y^2=37}_{x+y-1=0}\Leftrightarrow^{x^2-xy+y=37\left(1\right)}_{x+y=1\left(2\right)}\)
Nhân vế \(\left(1\right)\) với vế \(\left(2\right)\), ta có:
\(\left(x+y\right)\left(x^2-xy+y^2\right)=37.1\)
\(\Leftrightarrow x^3+y^3=37\)
\(\Leftrightarrow\left(x+y\right)^3-3xy\left(x+y\right)=37\)
\(\Leftrightarrow1-3xy=37\)
\(\Leftrightarrow3xy=-36\)
\(\Leftrightarrow xy=-12\)
Do đó: \(x^2-xy+y^2-xy=37-\left(-12\right)\)
\(\Leftrightarrow\left(x-y\right)^2=49\)
\(\Leftrightarrow x-y=7\) hoặc \(x-y=-7\)
Lại có: \(x+y=1\left(gt\right)\)
nên \(x=4;y=-3\) hoặc \(x=-3;y=4\)
Vậy, \(x,y\in\left\{\left(4;-3\right),\left(-3;4\right)\right\}\)
