Question

[Nhok baka]

Giải hệ phương trình:

\(\left\{{}\begin{matrix}xy+x+y=19\\x^2y+xy^2=84\end{matrix}\right.\)

Answer 1

[Nguyễn Ngọc Lộc]

Ta có : \(\left\{{}\begin{matrix}xy+x+y=19\left(I\right)\\x^2y+xy^2=84\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}xy+x+y=19\\xy\left(x+y\right)=84\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x+y=19-xy\\xy\left(19-xy\right)=84\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x+y=19-xy\\19xy-x^2y^2-84=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x+y=19-xy\\x^2y^2-12xy-7xy+84=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x+y=19-xy\\xy\left(xy-12\right)-7\left(xy-12\right)=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x+y=19-xy\\\left(xy-12\right)\left(xy-7\right)=0\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x+y=19-xy\\\left[{}\begin{matrix}xy-7=0\\xy-12=0\end{matrix}\right.\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}x+y=19-xy\\\left[{}\begin{matrix}xy=7\\xy=12\end{matrix}\right.\end{matrix}\right.\)

TH₁ : xy = 7 ( II )

=> \(x=\frac{7}{y}\)

- Thay xy = 7 ;\(x=\frac{7}{y}\) vào phương trình ( I ) ta được :

\(7+y+\frac{7}{y}=19\)

=> \(\frac{y^2}{y}+\frac{7}{y}=12\)

=> \(y^2-12y+7=0\)

=> \(y^2-2.y.6+36-29=0\)

=> \(\left(y-6\right)^2=29\)

=> \(\left[{}\begin{matrix}y-6=\sqrt{29}\\y-6=-\sqrt{29}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}y=6+\sqrt{29}\\y=6-\sqrt{29}\end{matrix}\right.\)

- Thay \(y=6+\sqrt{29};6-\sqrt{29}\) vào phương trình ( II ) ta được :

\(\left[{}\begin{matrix}x\left(6+\sqrt{29}\right)=7\\x\left(6-\sqrt{29}\right)=7\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{7}{6+\sqrt{29}}\\x=\frac{7}{6-\sqrt{29}}\end{matrix}\right.\)

TH₂ : xy = 12 ( III )

=> \(x=\frac{12}{y}\)

- Thay xy = 12 ;\(x=\frac{12}{y}\) vào phương trình ( I ) ta được :

\(12+y+\frac{12}{y}=19\)

=> \(\frac{y^2}{y}+\frac{12}{y}=7\)

=> \(y^2-7y+12=0\)

=> \(y^2-2.y.\frac{7}{2}+\frac{49}{4}-\frac{1}{4}=0\)

=> \(\left(y-\frac{7}{2}\right)^2=\frac{1}{4}\)

=> \(\left[{}\begin{matrix}y-\frac{7}{2}=\sqrt{\frac{1}{4}}\\y-\frac{7}{2}=-\sqrt{\frac{1}{4}}\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}y=\sqrt{\frac{1}{4}}+\frac{7}{2}=4\\y=\frac{7}{2}-\sqrt{\frac{1}{4}}=3\end{matrix}\right.\)

- Thay y=4 ; y=3 vào phương trình ( II ) ta được :

\(\left[{}\begin{matrix}x4=7\\x3=7\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x=\frac{7}{4}\\x=\frac{7}{3}\end{matrix}\right.\)

Vậy hệ phương trình có các nghiệm ( x; y ) là ( \(\frac{7}{4};4\) ) ; ( \(\frac{7}{3};3\) ) ;

( \(\frac{7}{6+\sqrt{29}};6+\sqrt{29}\) ) ; \(\left(\frac{7}{6-\sqrt{29}};6-\sqrt{29}\right)\)