\(\sqrt{x}=\sqrt{6+2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)
\(B=\frac{\sqrt{x}-1}{2+\sqrt{x}}=\frac{\sqrt{5}+1-1}{2+\sqrt{5}+1}=\frac{\sqrt{5}}{\sqrt{5}+3}=\frac{\left(3-\sqrt{5}\right)\sqrt{5}}{\left(3^2-5\right)}=\frac{3\sqrt{5}-5}{4}\)
\(B=\frac{\sqrt{x}-1}{2+\sqrt{x}}=\frac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\frac{3}{\sqrt{x}+2}\inℤ\Leftrightarrow\frac{3}{\sqrt{x}+2}\inℤ\)
mà \(x\)nguyên nên \(\sqrt{x}+2\inƯ\left(3\right)\)mà \(\sqrt{x}+2\ge2\)nên \(\sqrt{x}+2=3\Leftrightarrow x=1\).
