Thay\(x=\dfrac{16}{9}\)vào A ta đc
A=\(\dfrac{\sqrt{\dfrac{16}{9}}+1}{\sqrt{\dfrac{16}{9}}-1}=\dfrac{\dfrac{4}{3}+1}{\dfrac{4}{3}-1}=\dfrac{21}{3}=7\)
Thay x=\(\dfrac{25}{9}\)vào A ta đươc
A=\(\dfrac{\sqrt{\dfrac{25}{9}}+1}{\sqrt{\dfrac{25}{9}-1}}=\dfrac{\dfrac{5}{3}+1}{\dfrac{5}{3}-1}=\dfrac{8}{2}=4\)
Vậy....................