Với \(x\ge0;x\ne1\)

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

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

= -5, 17090592086

\(B=\left(3-\sqrt{5}\right)\sqrt{3+\sqrt{5}}+\left(3+\sqrt{5}\right)\sqrt{3+\sqrt{5}}\)

\(=\sqrt{3+\sqrt{5}}\left(3-\sqrt{5}+3+\sqrt{5}\right)\)

\(=6\sqrt{3+\sqrt{5}}\)

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

\(\Rightarrow B=\frac{6\left(\sqrt{5}+1\right)}{\sqrt{2}}=\frac{\sqrt{18.2}\left(\sqrt{5}+1\right)}{\sqrt{2}}=3\sqrt{2}\left(\sqrt{5}+1\right)=3\sqrt{10}+3\sqrt{2}\)

a, \(\left(2+\sqrt{5}\right)\sqrt{7-4\sqrt{3}}-\sqrt{\left(-3\right)^2}\)

\(=\left(2+\sqrt{5}\right)\sqrt{\left(2-\sqrt{3}\right)^2}-3\)

\(=\left(2+\sqrt{5}\right)\left(2-\sqrt{3}\right)-3=1-2\sqrt{3}+2\sqrt{5}-\sqrt{15}\)

b, đề sai ko bạn ? 

c, \(\frac{1}{2\sqrt{3}}-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-\sqrt{7-2\sqrt{6}}\)

\(=\frac{2\sqrt{3}}{12}-\frac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\sqrt{\left(\sqrt{6}-1\right)^2}\)

\(=\frac{\sqrt{3}}{6}-\sqrt{2}-\sqrt{6}+1=\frac{\sqrt{3}-6\sqrt{2}-6\sqrt{6}+6}{6}\)

zễ mà;-;

\(\frac{2}{3\sqrt{2}-4}-\frac{2}{3\sqrt{2}+4}\)

\(=\frac{6\sqrt{2}+8-6\sqrt{2}+8}{\left(3\sqrt{2}\right)^2-4^2}\)

\(=\frac{16}{18-16}=\frac{16}{2}=8\)