`B=(a+\sqrta+1)/(\sqrta-1)`
`B=[(a-2\sqrta+1)+(3\sqrta-3)+3]/(\sqrta-1)`
`B=[(\sqrta-1)^2+3(\sqrta-1)+3]/(\sqrta-1)`
`B=(\sqrta-1)+3/(\sqrta-1)+3`
`B≥2\sqrt[(\sqrta-1). 3/(\sqrta-1)]+3=2\sqrt3+3`
Dấu `=` xảy ra `⇔\sqrta-1=3/(\sqrta-1)⇒a=4+2\sqrt3`
Vậy $Min_B$`=2\sqrt3+3⇔a=4+2\sqrt3`