Question

Tìm x;y biết :

\(\int^{x^2-xy+y^2=37}_{x+y-1=0}\)

Answer 1

\(^{x^2-xy+y^2=37}_{x+y-1=0}\Leftrightarrow^{x^2-xy+y=37\left(1\right)}_{x+y=1\left(2\right)}\)

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,  \(x,y\in\left\{\left(4;-3\right),\left(-3;4\right)\right\}\)