[tex]x=ky[/tex]
[tex]\left\{\begin{matrix} \frac{6}{y(2k-3)}+\frac{2}{y(k+2)}=3(1)) & \\ \frac{3}{y(k-2)}+\frac{4}{y(k+2)}=-1(2) & \end{matrix}\right.[/tex]
[tex]\frac{(1)}{(2)}=\frac{\frac{6}{2k-3}+\frac{2}{k+2}}{\frac{3}{k-2}+\frac{4}{k+2}}=-3<=>\left ( k-\frac{89}{104} \right )^2=\frac{6673}{10816}[/tex]
=> [TEX]k[/TEX]
=> thay zô pt 1 hay 2 rồi giải