Question

rút gọn

B=\(\frac{1}{2-\sqrt{3}}-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-\sqrt{5-2\sqrt{6}}\)

C=\(\frac{15}{\sqrt{6}+1}-\frac{6}{\sqrt{6}-2}\)

Answer 1

\(B=\frac{1}{2-\sqrt{3}}-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-\sqrt{5-2\sqrt{6}}\)

=\(\frac{1}{2-\sqrt{3}}-\frac{\sqrt{2}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}\)

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

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

Answer 2

\(C=\frac{15}{\sqrt{6}+1}-\frac{6}{\sqrt{6}-2}=\frac{15\left(\sqrt{6}-1\right)}{6-1}-\frac{6\left(\sqrt{6}+2\right)}{6-4}\)

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

\(=-9\)

Answer 3

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