Giúp mình với ạ mình cần gấp

a) 

$\text{A = $\dfrac{1 + \sqrt{5}}{\sqrt{2} + \sqrt{3 + \sqrt{5}} }$} + $$\dfrac{1 - \sqrt{5}}{\sqrt{2} - \sqrt{3 - \sqrt{5}}}$

$\text{$\dfrac{A}{\sqrt{2}}$} = $ $\dfrac{1 + \sqrt{5}}{2 + \sqrt{6 + 2 \sqrt{5}} } + $ $\dfrac{1 - \sqrt{5}}{2 - \sqrt{6 - 2 \sqrt{5}}}$

$\text{A = $\dfrac{1 + \sqrt{5}}{2 + \sqrt{5} + 1} + $}$ $\dfrac{1 - \sqrt{5}}{2 - \sqrt{5} + 1}$

$\text{A = $\dfrac{1 + \sqrt{5}}{3 + \sqrt{5}}$} + $ $\dfrac{1 - \sqrt{5}}{3 - \sqrt{5}}$

$\text{A = $\dfrac{(1 + \sqrt{5})(3 - \sqrt{5})}{9 - 5}$} + $ $\dfrac{(1 - \sqrt{5})(3 + \sqrt{5})}{9 - 5}$

$\text{A= $\dfrac{3 + 2\sqrt{5} - 5 + 3 - 2\sqrt{5} - 5}{4}$}$

$\text{A = $-$1}$

Suy ra được: $\text{B = $-$$\sqrt{2}$}$

b) 

$\text{N$\sqrt{2}$ = $\sqrt{2x + 2\sqrt{2x - 1}}$} + $$\sqrt{2x - 2\sqrt{2x - 1}}$

$\text{N$\sqrt{2}$ = $\sqrt{(2x - 1) + 2\sqrt{2x - 1}}$} + 1 + $$\sqrt{2x - 1 - 2\sqrt{2x - 1 + 1}}$

$\text{N$\sqrt{2}$} = $$\sqrt{(\sqrt{2x - 1 + 1)^2}} + $$\sqrt{(2x - 1 - 1)^2}$

$\text{N$\sqrt{2}$} = |$$\sqrt{2x - 1 } + 1| + |1 - $$\sqrt{2x - 1}|$

$\text{N$\sqrt{2}$ $\ge$ |$\sqrt{2x - 1}$} + 1 + 1 - $$\sqrt{2x - 1}| $$\Rightarrow N$$\sqrt{2} $$\ge 2$ $\Rightarrow N \ge $$\sqrt{2}$

Dấu "$=$" xảy ra ⇔ $\sqrt{2x - 1} + 1 = 1- \sqrt{2x - 1}$

                            ⇒ $\text{2$\sqrt{2x - 1}$ = 0 ⇒ 2x - 1 ⇒ x = $\dfrac{1}{2}$}$

$\textit{#Ng}$