Răspuns:
Explicație pas cu pas:
[tex]\sqrt{12}=2\sqrt{3}\\\sqrt{27}=3\sqrt{3}\\\sqrt{108}=6\sqrt{3}\\=> x=(\frac{2}{2\sqrt{3} }+\frac{9}{3\sqrt{3} }+\frac{6}{6\sqrt{3} } ) \sqrt{3}\\x= (\frac{1+3+1}{\sqrt{3} })\sqrt{3}=5[/tex]
Deci x=5
