[tex]a(\sqrt5-2)+2b\sqrt5=3(\sqrt5-b)+1\\
a\sqrt5-2a+2b\sqrt5=3\sqrt5-3b+1\\
a\sqrt5+2b\sqrt5-3\sqrt5=2a-3b+1\\
\sqrt5(a+2b-3)=2a-3b+1\\
\sqrt5(a+b-3)\in \mathbb{Q}\Leftrightarrow a+2b-3=0\\
\text{Deci vom avea:}\\
\left \{ {{a+2b-3=0} \atop {2a-3b+1=0}} \right. \Leftrightarrow \left \{ {{a+2b=3} \atop {2a-3b=-1}} \right. \Leftrightarrow \left \{ {{2a+4b=6} \atop {2a-3b=-1}} \right. \\
\text{Scazand relatiile obtinem:}\\
/7b=7\Rightarrow \boxed{b=1}\\
a+2=3\\
\boxed{a=1}[/tex]
