Hệ phương trình x^3y(1+y)+x^2.y^2(2+y)+xy^3-30=0 và x^2y+x(1+y+y^2)+y-11=0

* Hướng dẫn giải

Hệ tương đương với $\left\{\begin{array}{l}xy\left(x+y\right)\left(x+y+xy\right)=30\\ xy\left(x+y\right)+x+y+xy=11\end{array}\right.$

Đặt xy (x + y) = a; xy + x + y = b. Ta thu được hệ:

$\left\{\begin{array}{l}ab=30\\ a+b=11\end{array}\right.\iff \left[\begin{array}{l}a=5;b=6\\ a=6;b=5\end{array}\right.\iff \left[\begin{array}{l}\left\{\begin{array}{l}xy\left(x+y\right)=5\\ xy+x+y=6\end{array}\right.\\ \left\{\begin{array}{l}xy\left(x+y\right)=6\\ xy+x+y=5\end{array}\right.\end{array}\right.$

TH1: $\left\{\begin{array}{l}xy\left(x+y\right)=6\\ xy+x+y=5\end{array}\right.$

$\iff \left[\begin{array}{l}\left\{\begin{array}{l}xy=2\\ x+y=3\end{array}\right.\\ \left\{\begin{array}{l}xy=3\\ x+y=2\end{array}\right.\left(L\right)\end{array}\right.\iff \left[\begin{array}{l}x=2;y=1\\ x=1;y=2\end{array}\right.$

TH2: $\left\{\begin{array}{l}xy\left(x+y\right)=5\\ xy+x+y=6\end{array}\right.$

$\iff \left[\begin{array}{l}\left\{\begin{array}{l}xy=5\\ x+y=1\end{array}\right.\left(L\right)\\ \left\{\begin{array}{l}xy=1\\ x+y=5\end{array}\right.\end{array}\right.\iff \left[\begin{array}{l}x=\frac{5-\sqrt{21}}{2};y=\frac{5+\sqrt{21}}{2}\\ x=\frac{5+\sqrt{21}}{2};y=\frac{5-\sqrt{21}}{2}\end{array}\right.$

Vậy hệ có nghiệm (x; y) = (1; 2), (2; 1), $\left(\frac{5\pm \sqrt{21}}{2};\frac{5\mp \sqrt{21}}{2}\right)$

Suy ra có một cặp nghiệm thỏa mãn đề bài là $\left(\frac{5-\sqrt{21}}{2};\frac{5+\sqrt{21}}{2}\right)$

Đáp án:D