Đáp án+Giải thích các bước giải:
$a.\\\dfrac{4}{\sqrt{5}-1}-\dfrac{6}{\sqrt{7}-1}\\= \dfrac{4(\sqrt{5}+1)}{(\sqrt{5}-1)(\sqrt{5}+1)}-\dfrac{6(\sqrt{7}+1)}{(\sqrt{7}-1)(\sqrt{7}+1)}\\= \sqrt{5}+1-\sqrt{7}-1\\= \sqrt{5}-\sqrt{7}\\b.\\\dfrac{3}{\sqrt{6}-\sqrt{3}}+\dfrac{4}{\sqrt{7}+\sqrt{3}}\\= \dfrac{3(\sqrt{6}+\sqrt{3})}{(\sqrt{6}-\sqrt{3})(\sqrt{6}+\sqrt{3})}+ \dfrac{4(\sqrt{7}-\sqrt{3})}{(\sqrt{7}+\sqrt{3})(\sqrt{7}-\sqrt{3})}\\= \sqrt{6}+\sqrt{3}+\sqrt{7}-\sqrt{3}\\= \sqrt{6}+\sqrt{7}\\c.\\\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\\= \sqrt{\dfrac{(2-\sqrt{3})(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}}+\sqrt{\dfrac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}}\\= \sqrt{(2-\sqrt{3})(2-\sqrt{3})} +\sqrt{(2+\sqrt{3})(2+\sqrt{3})}\\= \sqrt{(2-\sqrt{3})^2}+\sqrt{(2+\sqrt{3})^2}\\= 2-\sqrt{3} + 2+\sqrt{3}\\= 4\\d.\\\dfrac{1}{\sqrt{2}+\sqrt{2-\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\\= \dfrac{1(\sqrt{2} -\sqrt{2-\sqrt{3}})}{(\sqrt{2}+\sqrt{2-\sqrt{3}})(\sqrt{2}-\sqrt{2-\sqrt{3}})}+ \dfrac{1(\sqrt{2} +\sqrt{2-\sqrt{3}})}{(\sqrt{2}-\sqrt{2-\sqrt{3}})(\sqrt{2}+\sqrt{2-\sqrt{3}})}\\= \dfrac{\sqrt{2}-\sqrt{2-\sqrt{3}}}{\sqrt{3}}+\dfrac{\sqrt{2}+\sqrt{2-\sqrt{3}}}{\sqrt{3}}\\= \dfrac{\sqrt{2}-\sqrt{2-\sqrt{3}} +\sqrt{2} +\sqrt{2-\sqrt{3}}}{\sqrt{3}}\\= \dfrac{2\sqrt{2}}{\sqrt{3}}\\= \dfrac{2\sqrt{2} \times \sqrt{3}}{\sqrt{3}\times \sqrt{3}}\\= \dfrac{2\sqrt{6}}{3}$
