sửa đề : \(\sqrt{11-6\sqrt{2}}+3+\sqrt{2}\)

\(=\sqrt{9-2.3\sqrt{2}+2}+3+\sqrt{2}\)

\(=\sqrt{\left(3-\sqrt{2}\right)^2}+3+\sqrt{2}=3-\sqrt{2}+3+\sqrt{2}=6\)

Với \(x\ge0;x\ne1\)

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

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

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

1, \(\sqrt{12}+5\sqrt{3}-\sqrt{48}=2\sqrt{3}+5\sqrt{3}-4\sqrt{3}=3\sqrt{3}\)

2, \(5\sqrt{5}+\sqrt{20}-3\sqrt{45}=5\sqrt{5}+2\sqrt{5}-6\sqrt{5}=\sqrt{5}\)

còn đâu bạn tự làm, tương tự 

7, \(\sqrt{11+6\sqrt{2}}+\sqrt{11-6\sqrt{2}}=\sqrt{9+2.3\sqrt{2}+2}+\sqrt{9-2.3\sqrt{2}+2}\)

\(=\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}=3+\sqrt{2}+3-\sqrt{2}=6\)

8, \(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)

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

9, \(\sqrt{8-2\sqrt{7}}-\sqrt{11-4\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{7-4\sqrt{7}+4}\)

\(=\sqrt{7}-1-\sqrt{\left(\sqrt{7}-2\right)^2}=\sqrt{7}-1-\sqrt{7}+2=1\)