Chủ đề này có 2 trả lời

[Takamina Minami]

Rút gọn biểu thức: 

$C=\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$

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

$G=\sqrt{a+b+c+2\sqrt{ac+bc}}+\sqrt{a+b+c-2\sqrt{ac+bc}}$

[nghiemthanhbach]

$G=\sqrt{a+b+c+2\sqrt{ac+bc}}+\sqrt{a+b+c-2\sqrt{ac+bc}}=\sqrt{\frac{a+b+c+\sqrt{(a+b+c)^2-4ac-4bc}}{2}}+\sqrt{\frac{a+b+c-\sqrt{(a+b+c)^2-4ac-4bc}}{2}}+\sqrt{\frac{a+b+c+\sqrt{(a+b+c)^2-4ac-4bc}}{2}}-\sqrt{\frac{a+b+c-\sqrt{(a+b+c)^2-4ac-4bc}}{2}}=\sqrt{2[a+b+c+\sqrt{(a+b-c)^2}]}=\sqrt{2(a+b+c+|a+b-c)}=\begin{bmatrix} 2\sqrt{a+b}\\ 2\sqrt{c} \end{bmatrix}$

[nghiemthanhbach]

$\sqrt{x\pm \sqrt{y}}=\sqrt{\frac{x+\sqrt{x^2-y}}{2}}\pm \sqrt{\frac{x-\sqrt{x^2-y}}{2}}$