Đáp án:
Giải thích các bước giải:
$B=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}$
$\dfrac{1}{\sqrt{2}}B=\dfrac{2+\sqrt{3}}{\sqrt{2}(\sqrt{2}+\sqrt{2+\sqrt{3}})}+\dfrac{2-\sqrt{3}}{\sqrt{2}(\sqrt{2}-\sqrt{2-\sqrt{3}})}$
$\dfrac{1}{\sqrt{2}}B=\dfrac{2+\sqrt{3}}{2+\sqrt{3}+1}+\dfrac{2-\sqrt{3}}{2-\sqrt{3}+1}$
$\dfrac{1}{\sqrt{2}}B=\dfrac{2+\sqrt{3}}{3+\sqrt{3}}+\dfrac{2-\sqrt{3}}{3-\sqrt{3}}$
$\dfrac{1}{\sqrt{2}}B=\dfrac{(2+\sqrt{3})(3-\sqrt{3})}{9-3}+\dfrac{(2-\sqrt{3})(3+\sqrt{3})}{9-3}$
$\dfrac{1}{\sqrt{2}}B=\dfrac{6+\sqrt{3}-3+6-\sqrt{3}-3}{6}$
$\dfrac{1}{\sqrt{2}}B=\dfrac{12-6}{6}=1$
$⇒B=\sqrt{2}$
