[tex]k)~( \sqrt{2} + \sqrt{3} )x- \sqrt{8} = \sqrt{12}=\ \textgreater \ x= \frac{ \sqrt{12}+ \sqrt{8} }{ \sqrt{2}+ \sqrt{3} }= \frac{ \sqrt{4}( \sqrt{3}+ \sqrt{2}) }{ \sqrt{3}+ \sqrt{2} }= \sqrt{4}=2. \\ \boxed{Solutie:~x=2}.\\ \\ l)~ \sqrt{7} x-3 \sqrt{7} =9-x\ \textless \ =\ \textgreater \ \sqrt{7}x+x=9+3 \sqrt{7} \ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ x( \sqrt{7}+1)= 3( 3+\sqrt{7})=\ \textgreater \ x= \frac{3(3+ \sqrt{7}) }{ \sqrt{7}+1 } = \frac{3(3+ \sqrt{7})( \sqrt{7}-1) }{7-1}= \\ = \frac{3( 3\sqrt{7}-3+7- \sqrt{7} )}{6}= \frac{ 2\sqrt{7} +4}{2} = \sqrt{7}+2.[/tex]
[tex]\boxed{Solutie:~x= \sqrt{7}+2}. [/tex]
[tex]m)~ \sqrt{8}x- \sqrt{2}x- \sqrt{10}= \sqrt{40} \ \textless \ =\ \textgreater \ \sqrt{2} (2x-x- \sqrt{5})= \sqrt{2}* 2\sqrt{5}\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ x- \sqrt{5}= 2\sqrt{5} =\ \textgreater \ x= 3\sqrt{5}. \\ \boxed{Solutie:~x= 3\sqrt{5}} . \\ \\ n)~(1- \sqrt{3})x- 4\sqrt{5}=(2- \sqrt{3})x+ \sqrt{45}\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ (1- \sqrt{3})x-(2- \sqrt{3})x= \sqrt{45}+ 4\sqrt{5}\ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ x[(1- \sqrt{3})-(2- \sqrt{3})]=3 \sqrt{5}+ 4\sqrt{5} \ \textless \ =\ \textgreater \ \\ \ \textless \ =\ \textgreater \ -x= 7\sqrt{5}=\ \textgreater \ x=- 7\sqrt{5}. \\ \boxed{Solutie:~x=-7\sqrt{5}} .[/tex]
