giải hệ pt a, x^2y+xy^2=6 và xy+x+y=5 b, x-y+xy=-1 và x^2+y^2-x-y=2

a) $\left \{ {{xy(x+y)=6} \atop {xy+(x+y)=5}} \right.$ 

Đặt: xy=a; x+y=b

Ta có: $\left \{ {{ab=6} \atop {a+b=5}} \right.$ ⇔$\left \{ {{(5-b)b=6} \atop {a=5-b}} \right.$ ⇔\(\left[ \begin{array}{l}a=2;b=3\\a=3;b=2\end{array} \right.\) 

TH1: $\left \{ {{xy=2} \atop {x+y=3}} \right.$ ⇔\(\left[ \begin{array}{l}x=1;y=2\\x=2;y=1\end{array} \right.\) 

TH2: $\left \{ {{xy=3} \atop {x+y=2}} \right.$ ⇔$\left \{ {{(2-y)y=3} \atop {x=2-y}} \right.$ ⇔\(\left[ \begin{array}{l}x=3;y=-1\\x=-1;y=3\end{array} \right.\)