[tex]1)~2 \sqrt{3}x=2 \sqrt{27} =\ \textgreater \ x= \frac{ 2\sqrt{27} }{2 \sqrt{3} }\ \textless \ =\ \textgreater \ \boxed{x=3}. \\ 2)~ \sqrt{(x-2) ^{2} } \leq 3\ \textless \ =\ \textgreater \ |x-2| \leq 3 =\ \textgreater \ -3 \leq x-2 \leq 3=\ \textgreater \ \\ =\ \textgreater \ -1 \leq x \leq 5\ \textless \ =\ \textgreater \ \boxed{x\in[-1;5]}. \\ 3)~f( \sqrt{3})=5\ \textless \ =\ \textgreater \ \sqrt{3}* \sqrt{3}+a=5\ \textless \ =\ \textgreater \ 3+a=5=\ \textgreater \ \boxed{a=2}. \\ 4)~ \left \{ {{2x+y=0} \atop {3x+2y=-2}} \right.\ \textless \ =\ \textgreater \ \left \{ {{4x+2y=0} \atop {3x+2y=-2}} \right. \ \textless \ =\ \textgreater \ \left \{ {{3x+2y=-2} \atop {x=2}} \right. \ \textless \ =\ \textgreater \ \left \{ {{y= \frac{-2-3x}{2} } \atop {x=2}} \right. [/tex]
[tex]\ \textless \ =\ \textgreater \ \left \{ {{y=-4} \atop {x=2}} \right. \\ Solutie:~\boxed{(x;y)=(2;-4)}.[/tex]
5) Pentru a gasi intersectia graficului functiei cu axa ordonatelor, calculam f(0).
[tex]f(0)=-3*0+1=1. \\ G_{f}\bigcap Oy=\boxed{A(0;1)}. [/tex]
[tex]6)~ \frac{x+1}{3}-1= \frac{x}{6} ~~~~~~~~~~~|*6 \\ 2(x+1)-6=x \\ 2x+2-6=x \\ 2x-x=6-2 \\ \boxed{x=4}.[/tex]
