Question

se considera nuerele reale ×=2√3si=8√3 acslculati 1 supra ×-1supra y

Answer 1

Din cate am inteles din enunt 
x=2√3
y=8√3
$\frac{1}{\frac{x-1}{y} }$ = $\frac{y}{x-1} = \frac{8\sqrt{3}}{2\sqrt{3}-1} = \frac{4*2\sqrt{3}}{2\sqrt{3}-1}= \frac{4*2\sqrt{3}*(2\sqrt{3}-1)}{12-1} = \frac{48-8\sqrt{3}}{11}= \frac{8*(6-\sqrt{3})}{11}$