Răspuns:
Explicație pas cu pas:
[tex]a=-\frac{1}{2}+\frac{1}{3} :\frac{1}{4}\\a= -\frac{1}{2}+\frac{1}{3}*\frac{4}{1}\\a= -\frac{1}{2}+\frac{4}{3}=\frac{-3+8}{6}=\frac{5}{6}[/tex] Adevarat
N=[tex]2a-5b[/tex]
N=[tex]2*\frac{5}{6}-5((\frac{\sqrt{3} }{2}-\frac{2}{\sqrt{3} } )^2-(\frac{\sqrt{3} }{2}-1)(1+\frac{\sqrt{3} }{2})=\\=\frac{5}{3}-5(\frac{3}{4}-2*\frac{\sqrt{3} }{2}*\frac{2}{\sqrt{3} }+\frac{4 }{3})-((\frac{\sqrt{3} }{2})^2+1)=\\=\frac{5}{3}-\frac{15}{4}+10-\frac{20}{3} -\frac{3}{4} -1=\\=-\frac{15}{3}-\frac{12}{4}+9=-5-3+9=1[/tex]
∈N
