Question

[dam thu a]

cho hpt \(\left\{{}\begin{matrix}mx+4y=9\\x+my=8\end{matrix}\right.\) . Tìm m để hệ có nghiệm (x,y) thỏa mãn \(2x+y+\frac{38}{m^2-4}=3\)

Answer 1

[Nguyễn Ngọc Lộc]

Ta có : \(\left\{{}\begin{matrix}mx+4y=9\\x+my=8\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}m\left(8-my\right)+4y=9\\x=8-my\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}8m-m^2y+4y=9\\x=8-my\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y\left(4-m^2\right)=9-8m\\x=8-my\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}y=\frac{9-8m}{4-m^2}\\x=8-\frac{m\left(9-8m\right)}{4-m^2}\end{matrix}\right.\)

- Ta có : \(2x+y+\frac{38}{m^2-4}=0\)

- Thay \(x=8-\frac{m\left(9-8m\right)}{4-m^2},y=\frac{9-8m}{4-m^2}\) vào phương trình trên ta được :

\(2\left(8-\frac{m\left(9-8m\right)}{4-m^2}\right)+\frac{9-8m}{4-m^2}+\frac{38}{m^2-4}=3\)

=> \(16-\frac{2m\left(9-8m\right)}{4-m^2}+\frac{9-8m}{4-m^2}-\frac{38}{4-m^2}=3\)

=> \(\frac{2m\left(9-8m\right)}{4-m^2}-\frac{9-8m}{4-m^2}+\frac{38}{4-m^2}=13\)

=> \(\frac{18m-16m^2-9+8m+38}{4-m^2}=13\)

=> \(26m-16m^2+29=13\left(4-m^2\right)\)

=> \(26m-16m^2+29-52+13m^2=0\)

=> \(3m^2-26m+23=0\)

=> \(\left(3m-23\right)\left(m-1\right)=0\)

=> \(\left[{}\begin{matrix}3m-23=0\\m-1=0\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}m=\frac{23}{3}\\m=1\end{matrix}\right.\)

Vậy m = 23/3, m = 1 thỏa mãn điều kiện trên .