Question

Cho A = \(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\). Chứng minh rằng với x = \(\dfrac{16}{9}\) và x = \(\dfrac{25}{9}\) thì A có giá trị nguyên

Answer 1

\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)

Với \(x=\dfrac{16}{9}\)

\(A=\dfrac{\sqrt{\dfrac{16}{9}}+1}{\sqrt{\dfrac{16}{9}}-1}\)

\(A=\dfrac{\dfrac{4}{3}+1}{\dfrac{4}{3}-1}=\dfrac{\dfrac{7}{3}}{\dfrac{-1}{3}}=7:3:-1.3=-7\)

Với \(x=\dfrac{25}{9}\)

\(A=\dfrac{\sqrt{\dfrac{25}{9}}+1}{\sqrt{\dfrac{25}{9}}-1}\)

\(A=\dfrac{\dfrac{5}{3}+1}{\dfrac{5}{3}-1}=\dfrac{\dfrac{8}{3}}{\dfrac{2}{3}}=8:3:2.3=4\)

\(\rightarrowđpcm\)