Câu hỏi của Như Huỳnh

Question

Tính

$\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}$

\(\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\d...

qwerty · Accepted Answer

a) $\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}$

$=\dfrac{\left(245-100\sqrt{6}+98\sqrt{6}-240\right)\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}}{9\sqrt{3}-11\sqrt{2}}$

$=\dfrac{\left(5-2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)}{9\sqrt{3}-11\sqrt{2}}$

$=\dfrac{5\sqrt{3}-5\sqrt{2}-2\sqrt{18}+2\sqrt{12}}{9\sqrt{3}-11\sqrt{2}}$

$=\dfrac{5\sqrt{3}-5\sqrt{2}-6\sqrt{2}+4\sqrt{3}}{9\sqrt{3}-11\sqrt{2}}$

$=\dfrac{9\sqrt{3}-11\sqrt{2}}{9\sqrt{3}-11\sqrt{2}}$

$=1$Câu trả lời: a) $\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}$

$=\dfrac{\left(245-100\sqrt{6}+98\sqrt{6}-240\right)\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}}{9\sqrt{3}-11\sqrt{2}}$

$=\dfrac{\left(5-2\sqrt{6}\right)\left(\sqrt{3}-\sqrt{2}\right)}{9\sqrt{3}-11\sqrt{2}}$

$=\dfrac{5\sqrt{3}-5\sqrt{2}-2\sqrt{18}+2\sqrt{12}}{9\sqrt{3}-11\sqrt{2}}$

$=\dfrac{5\sqrt{3}-5\sqrt{2}-6\sqrt{2}+4\sqrt{3}}{9\sqrt{3}-11\sqrt{2}}$

$=\dfrac{9\sqrt{3}-11\sqrt{2}}{9\sqrt{3}-11\sqrt{2}}$

$=1$

qwerty · Answer

b)

$\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}$

$=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2\sqrt{6}}{6}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}$

$=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{\sqrt{6}}{3}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}}$

$=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{3\sqrt{3\left(2+\sqrt{3}\right)}-2\sqrt{18}+3\sqrt{2+\sqrt{3}}}{6\sqrt{3}}}$

$=\dfrac{\dfrac{\sqrt{2+\sqrt{3}}}{2}}{\dfrac{3\sqrt{6+3\sqrt{3}-6\sqrt{2}+3\sqrt{2+\sqrt{3}}}}{6\sqrt{3}}}$

$=\dfrac{3\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{3\sqrt{6+3\sqrt{3}}-6\sqrt{2}+3\sqrt{2+\sqrt{3}}}$

$=\dfrac{3\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{3\left(\sqrt{6+3\sqrt{3}}-2\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}$

$=\dfrac{\sqrt{\left(2+\sqrt{3}\right)\cdot3}}{\sqrt{6+3\sqrt{3}}-2\sqrt{2}+\sqrt{2+\sqrt{3}}}$

$=\dfrac{\sqrt{6+3\sqrt{3}}}{\sqrt{6+3\sqrt{3}}-2\sqrt{2}+\sqrt{2+\sqrt{3}}}$

$=\dfrac{\sqrt{\left(6+3\sqrt{3}\right)\left(-\sqrt{3}+2+\sqrt{3}\right)}}{-2\sqrt{3}}$

$=\dfrac{\sqrt{\left(6+3\sqrt{3}\right)\cdot2}}{-2\sqrt{3}}$

$=\dfrac{\sqrt{12+6\sqrt{3}}}{-2\sqrt{3}}$

$=\dfrac{\sqrt{\left(3+\sqrt{3}\right)^2}}{-2\sqrt{3}}$

$=\dfrac{3+\sqrt{3}}{-2\sqrt{3}}$

$=-\dfrac{\left(3+\sqrt{3}\right)\sqrt{3}}{6}$

$=-\dfrac{3\sqrt{3}+3}{6}$

$=-\dfrac{3\left(\sqrt{3}+3\right)}{6}$

$=-\dfrac{\sqrt{3}+1}{2}$Câu trả lời: b)