$\frac{1}{2-\sqrt{3}}-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-\sqrt{5-2\sqrt{6}}_{}$ 01/08/2021 Bởi Autumn $\frac{1}{2-\sqrt{3}}-\frac{\sqrt{6}-\sqrt{2}}{\sqrt{3}-1}-\sqrt{5-2\sqrt{6}}_{}$
$\begin{array}{l}\dfrac{1}{2 – \sqrt3} – \dfrac{\sqrt6 – \sqrt2}{\sqrt3 – 1} – \sqrt{5 – 2\sqrt6}\\ = \dfrac{2 + \sqrt3}{(2 – \sqrt3)(2 + \sqrt3)} – \dfrac{\sqrt2(\sqrt3 – 1)}{\sqrt3 – 1} – \sqrt{3 – 2\sqrt3.\sqrt2 + 2}\\ = \dfrac{2 + \sqrt3}{4 – 3} – \sqrt2 – \sqrt{(\sqrt3 – \sqrt2)^2}\\ = 2 + \sqrt3 – \sqrt2 – (\sqrt3 – \sqrt2)\\ = 2\end{array}$ Bình luận
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$\begin{array}{l}\dfrac{1}{2 – \sqrt3} – \dfrac{\sqrt6 – \sqrt2}{\sqrt3 – 1} – \sqrt{5 – 2\sqrt6}\\ = \dfrac{2 + \sqrt3}{(2 – \sqrt3)(2 + \sqrt3)} – \dfrac{\sqrt2(\sqrt3 – 1)}{\sqrt3 – 1} – \sqrt{3 – 2\sqrt3.\sqrt2 + 2}\\ = \dfrac{2 + \sqrt3}{4 – 3} – \sqrt2 – \sqrt{(\sqrt3 – \sqrt2)^2}\\ = 2 + \sqrt3 – \sqrt2 – (\sqrt3 – \sqrt2)\\ = 2\end{array}$