Chủ đề này có 1 trả lời

[Takamina Minami]

rút gọn biểu thức            
$C=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$

$E=\frac{1+\frac{\sqrt{3}}{2}}{1+\sqrt{1}+\frac{\sqrt{3}}{2}}+\frac{1-\frac{\sqrt{3}}{2}}{1-\sqrt{\frac{\sqrt{3}}{2}}}$

[CHU HOANG TRUNG]

rút gọn biểu thức            
$C=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$

 

$C\sqrt{2}=\sqrt{2}(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}})=\frac{\sqrt{2}.\sqrt{2}.(2+\sqrt{3})}{\sqrt{2}(\sqrt{2}+\sqrt{2+\sqrt{3}})}+\frac{\sqrt{2}.\sqrt{2}.(2-\sqrt{3})}{\sqrt{2}(\sqrt{2}-\sqrt{2-\sqrt{3}})}=\frac{4+2\sqrt{3}}{2+\sqrt{4+2\sqrt{3}}}+\frac{4-2\sqrt{3}}{2-\sqrt{4-2\sqrt{3}}}=\frac{(\sqrt{3}+1)^2}{2+\sqrt{(\sqrt{3}+1)^2}}+\frac{(\sqrt{3}-1)^2}{2-\sqrt{(\sqrt{3}-1)^2}}=\frac{(\sqrt{3}+1)^2}{3+\sqrt{3}}+\frac{(\sqrt{3}-1)^2}{3-\sqrt{3}}=\frac{(\sqrt{3}+1)^2}{\sqrt{3}(\sqrt{3}+1)}+\frac{(\sqrt{3}-1)^2}{\sqrt{3}(\sqrt{3}-1)}=\frac{\sqrt{3}+1}{\sqrt{3}}+\frac{\sqrt{3}-1}{\sqrt{3}}=\frac{2\sqrt{3}}{\sqrt{3}}=2$

$\Rightarrow C\sqrt{2}=2\Rightarrow C=\sqrt{2}$