a: \(x=6+2\sqrt5=\left(\sqrt5+1\right)^2\)

\(\sqrt{x}-1\)

\(=\sqrt{\left(\sqrt5+1\right)^2}-1=\sqrt5+1-1=\sqrt5\)

\(2x+2-2x\cdot\sqrt{x}\)

\(=2\left(6+2\sqrt5\right)+2-2\cdot\left(6+2\sqrt5\right)\sqrt{\left(\sqrt5+1\right)^2}\)

\(=12+4\sqrt5+2-2\left(6+2\sqrt5\right)\left(\sqrt5+1\right)\)

\(=14+4\sqrt5-2\left(6\sqrt5+6+10+2\sqrt5\right)=14+4\sqrt5-2\left(8\sqrt5+16\right)\)

\(=14+4\sqrt5-16\sqrt5-32=-18-12\sqrt5\)

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

\(=\frac{\sqrt5}{-18-12\sqrt5}=\frac{-\sqrt5}{18+12\sqrt5}=\frac{-\sqrt5}{6\left(3+2\sqrt5\right)}=\frac{-\sqrt5\left(2\sqrt5-3\right)}{6\left(2\sqrt5+3\right)\left(2\sqrt5-3\right)}\)

\(=\frac{-10+3\sqrt5}{6\cdot\left(20-9\right)}=\frac{-10+3\sqrt5}{66}\)

a: \(x=6+2\sqrt5=\left(\sqrt5+1\right)^2\)

\(\sqrt{x}-1\)

\(=\sqrt{\left(\sqrt5+1\right)^2}-1=\sqrt5+1-1=\sqrt5\)

\(2x+2-2x\cdot\sqrt{x}\)

\(=2\left(6+2\sqrt5\right)+2-2\cdot\left(6+2\sqrt5\right)\sqrt{\left(\sqrt5+1\right)^2}\)

\(=12+4\sqrt5+2-2\left(6+2\sqrt5\right)\left(\sqrt5+1\right)\)

\(=14+4\sqrt5-2\left(6\sqrt5+6+10+2\sqrt5\right)=14+4\sqrt5-2\left(8\sqrt5+16\right)\)

\(=14+4\sqrt5-16\sqrt5-32=-18-12\sqrt5\)

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

\(=\frac{\sqrt5}{-18-12\sqrt5}=\frac{-\sqrt5}{18+12\sqrt5}=\frac{-\sqrt5}{6\left(3+2\sqrt5\right)}=\frac{-\sqrt5\left(2\sqrt5-3\right)}{6\left(2\sqrt5+3\right)\left(2\sqrt5-3\right)}\)

\(=\frac{-10+3\sqrt5}{6\cdot\left(20-9\right)}=\frac{-10+3\sqrt5}{66}\)

a.

\(x=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\) \(\Rightarrow\sqrt{x}=\sqrt{5}+1\)

 \(\Rightarrow B=\dfrac{\sqrt{5}+1-1}{2+\sqrt{5}+1}=\dfrac{\sqrt{5}}{\sqrt{5}+3}=\dfrac{3\sqrt{5}-5}{4}\)

b.

\(B=\dfrac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\dfrac{3}{\sqrt{x}+2}\)

B nguyên \(\Rightarrow\dfrac{3}{\sqrt{x}+2}\in Z\Rightarrow\sqrt{x}+2=Ư\left(3\right)\)

Mà \(\sqrt{x}+2\ge2\Rightarrow\sqrt{x}+2=3\)

\(\Leftrightarrow\sqrt{x}=1\Rightarrow x=1\)

Mình cảm ơn.

\(\sqrt{x}=\sqrt{6+2\sqrt{5}}=\sqrt{5+2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)

\(B=\frac{\sqrt{x}-1}{2+\sqrt{x}}=\frac{\sqrt{5}+1-1}{2+\sqrt{5}+1}=\frac{\sqrt{5}}{\sqrt{5}+3}=\frac{\left(3-\sqrt{5}\right)\sqrt{5}}{\left(3^2-5\right)}=\frac{3\sqrt{5}-5}{4}\)

\(B=\frac{\sqrt{x}-1}{2+\sqrt{x}}=\frac{\sqrt{x}+2-3}{\sqrt{x}+2}=1-\frac{3}{\sqrt{x}+2}\inℤ\Leftrightarrow\frac{3}{\sqrt{x}+2}\inℤ\)

mà \(x\)nguyên nên \(\sqrt{x}+2\inƯ\left(3\right)\)mà \(\sqrt{x}+2\ge2\)nên \(\sqrt{x}+2=3\Leftrightarrow x=1\).