e)
+ Ta có: $\frac{x}{y}$ = $\frac{2}{3}$ => x = 2.$\frac{y}{3}$
+ Ta có: x + y - 10 = 0 => $\frac{2y}{3}$ + y - 10 = 0 => $\frac{5y}{3}$ = 10 => \(\left[ \begin{array}{l}x=6\\y=9\end{array} \right.\) .
h)
+ Ta có: $\left \{ {{(1 + \sqrt{2})x + (1 + \sqrt{2})y - (1 + \sqrt{2})x -(1-\sqrt{2})y= -2} \atop {(1+\sqrt{2})x+(1+\sqrt{2})y=3}} \right.$
<=> $\left \{ {{(1+\sqrt{2})y -(1-\sqrt{2})y=-2} \atop {(1+\sqrt{2})(x+y)=3}} \right.$
<=> $\left \{ {{(1+\sqrt{2} -(1+\sqrt{2})y=-2} \atop {x+y=\frac{3}{1+\sqrt{2}}}} \right.$
<=> $\left \{ {{y=2\sqrt{2}y=-2}\atop {x+y=\frac{3.(\sqrt{2}-1)}{(1+\sqrt{2}).(\sqrt{2}-1)}}}\right.$
<=> $\left \{ {{y=\frac{-\sqrt{2}}{2} } \atop {x+y=\frac{3.(\sqrt{2}-1)}{2-1}}} \right.$
<=> $\left \{ {{y=\frac{-\sqrt{2}}{2} } \atop {x+y=3\sqrt{2}-3}} \right.$
<=> $\left \{ {{y=\frac{-\sqrt{2}}{2} } \atop {x+y=3\sqrt{2}-3-\frac{-\sqrt{2}}{2} }} \right.$
<=> $\left \{ {{y=\frac{-\sqrt{2}}{2} } \atop {x+y=3\sqrt{2}-3+\frac{\sqrt{2}}{2} }} \right.$
<=> $\left \{ {{y=\frac{-\sqrt{2}}{2} } \atop {x=\frac{6\sqrt{2}-6+\sqrt{2}}{2}=\frac{7\sqrt{2}-6}{2}}} \right.$.
+ Vậy hệ phương trình có nghiệm duy nhất ($\frac{7\sqrt{2}-6}{2}$;$\frac{-\sqrt{2}}{2}$).
