giải hệ phương trình: $\left \{ {{x+y+xy=7} \atop {x^3+y^3+3(x^2+y^2)+3(x+y)=70}} \right.$ 02/12/2021 Bởi Jasmine giải hệ phương trình: $\left \{ {{x+y+xy=7} \atop {x^3+y^3+3(x^2+y^2)+3(x+y)=70}} \right.$
$\left\{\begin{array}{l} x+y+xy=7\\ x^3+y^3+3(x^2+y^2)+3(x+y)=70\end{array} \right. <=>\left\{\begin{array}{l} xy=7-x-y\\ (x+y)(x^2-xy+y^2)+3(x^2+y^2)+3(x+y)=70\end{array} \right.\\ <=>\left\{\begin{array}{l} xy=7-x-y\\ (x+y)((x+y)^2-3xy)+3((x+y)^2-2xy)+3(x+y)=70\end{array} \right.\\ <=>\left\{\begin{array}{l} xy=7-x-y\\ (x+y)[(x+y)^2-21+3(x+y)]+3(x+y)^2-42+6(x+y)+3(x+y)=70\end{array} \right.\\ <=>\left\{\begin{array}{l} xy=7-x-y\\ (x+y)^3-21(x+y)+3(x+y)^2+3(x+y)^2-42+6(x+y)+3(x+y)=70\end{array} \right.\\ <=>\left\{\begin{array}{l} xy=7-x-y\\ (x+y)^3+6(x+y)^2-12(x+y)-112=0\end{array} \right.\\ <=>\left\{\begin{array}{l} xy=7-x-y\\ x+y=4\end{array} \right.\\ <=>\left\{\begin{array}{l} xy=3\\ x+y=4\end{array} \right.\\$$ <=>\left[\begin{array}{l} \left\{\begin{array}{l} x=1\\ y=3\end{array} \right.\\ \left\{\begin{array}{l} x=3\\ y=1\end{array} \right.\end{array} \right.$ Bình luận
$\left\{\begin{array}{l} x+y+xy=7\\ x^3+y^3+3(x^2+y^2)+3(x+y)=70\end{array} \right. <=>\left\{\begin{array}{l} xy=7-x-y\\ (x+y)(x^2-xy+y^2)+3(x^2+y^2)+3(x+y)=70\end{array} \right.\\ <=>\left\{\begin{array}{l} xy=7-x-y\\ (x+y)((x+y)^2-3xy)+3((x+y)^2-2xy)+3(x+y)=70\end{array} \right.\\ <=>\left\{\begin{array}{l} xy=7-x-y\\ (x+y)[(x+y)^2-21+3(x+y)]+3(x+y)^2-42+6(x+y)+3(x+y)=70\end{array} \right.\\ <=>\left\{\begin{array}{l} xy=7-x-y\\ (x+y)^3-21(x+y)+3(x+y)^2+3(x+y)^2-42+6(x+y)+3(x+y)=70\end{array} \right.\\ <=>\left\{\begin{array}{l} xy=7-x-y\\ (x+y)^3+6(x+y)^2-12(x+y)-112=0\end{array} \right.\\ <=>\left\{\begin{array}{l} xy=7-x-y\\ x+y=4\end{array} \right.\\ <=>\left\{\begin{array}{l} xy=3\\ x+y=4\end{array} \right.\\$$ <=>\left[\begin{array}{l} \left\{\begin{array}{l} x=1\\ y=3\end{array} \right.\\ \left\{\begin{array}{l} x=3\\ y=1\end{array} \right.\end{array} \right.$