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[The Flash]

Tìm $a,b\in \mathbb{Q}$ sao cho $\sqrt{a}+\sqrt{b}=\sqrt{2+\sqrt{3}}$

[L Lawliet]

Tìm $a,b\in \mathbb{Q}$ sao cho $\sqrt{a}+\sqrt{b}=\sqrt{2+\sqrt{3}}$

Lời giải.

Ta có:

\begin{align*} \sqrt{2+\sqrt{3}} &=\dfrac{\sqrt{2}}{2}\sqrt{4+2\sqrt{3}} \\ &=\dfrac{\sqrt{2}}{2}\sqrt{\left ( \sqrt{3}+1 \right )^{2}} \\ &=\dfrac{\sqrt{2}}{2}\left | \sqrt{3}+1 \right | \\ &=\dfrac{\sqrt{2}}{2}\left ( \sqrt{3}+1 \right ) \\ &=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{2}}{2} \end{align*}

$$\Rightarrow \sqrt{a}+\sqrt{b}=\dfrac{\sqrt{6}}{2}+\dfrac{\sqrt{2}}{2}$$

Vậy $a=\dfrac{3}{2}$ và $b=\dfrac{1}{2}$ hoặc $a=\dfrac{1}{2}$ và $b=\dfrac{3}{2}$.