Question

\(B=\dfrac{5\left(\sqrt{6}-1\right)}{\sqrt{6}+1}+\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}+\sqrt{3}}+\sqrt{3-2\sqrt{2}}\)

Answer 1

Lời giải:

\(\frac{5(\sqrt{6}-1)}{\sqrt{6}+1}=\frac{5(\sqrt{6}-1)^2}{(\sqrt{6}+1)(\sqrt{6}+1)}=\frac{5(\sqrt{6}-1)^2}{6-1}=(\sqrt{6}-1)^2\)

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

\(\sqrt{3-2\sqrt{2}}=\sqrt{\frac{6-2\sqrt{8}}{2}}=\sqrt{\frac{4+2-2\sqrt{2.4}}{2}}=\sqrt{\frac{(2-\sqrt{2})^2}{2}}=\frac{2-\sqrt{2}}{\sqrt{2}}=\sqrt{2}-1\)

Do đó:

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