Question

Answer 1

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

\(=\dfrac{2\sqrt{x}+1}{x-1}\cdot\dfrac{\sqrt{x}-1}{1}=\dfrac{2\sqrt{x}+1}{\sqrt{x}+1}\)

b: \(M=1+\dfrac{x-\sqrt{x}-5}{\sqrt{x}+3}\)

\(=\dfrac{\sqrt{x}+3+x-\sqrt{x}-5}{\sqrt{x}+3}\)

\(=\dfrac{x-2}{\sqrt{x}+3}\)

Để M là số nguyên thì \(x+3\sqrt{x}-3\sqrt{x}-9+7⋮\sqrt{x}+3\)

\(\Leftrightarrow\sqrt{x}+3=7\)

hay x=16