Odpowiedź:
[tex]\huge\boxed{D.\ \dfrac{\sqrt3-1}{\sqrt3+1}}[/tex]
Szczegółowe wyjaśnienie:
[tex]\dfrac{\sqrt3-\dfrac{1}{\sqrt{16}}:\dfrac{1}{4}\cdot\sqrt{\dfrac{36}{49}}\cdot\left(1\dfrac{1}{6}\right)}{\sqrt3+\sqrt{1\dfrac{7}{9}}\cdot\sqrt[3]{27}\cdot4^0:4}=\dfrac{\sqrt3-\dfrac{1}{4\!\!\!\!\diagup}\cdot\dfrac{4\!\!\!\!\diagup^1}{1}\cdot\dfrac{6\!\!\!\!\diagup}{7\!\!\!\!\diagup}\cdot\dfrac{7\!\!\!\!\diagup}{6\!\!\!\!\diagup}}{\sqrt3+\sqrt{\dfrac{16}{9}}\cdot3\cdot1\cdot\dfrac{1}{4}}=\dfrac{\sqrt3-1}{\sqrt3+\dfrac{4\!\!\!\!\diagup}{3\!\!\!\!\diagup}\cdot3\!\!\!\!\diagup\cdot\dfrac{1}{4\!\!\!\!\diagup}}[/tex]
[tex]=\dfrac{\sqrt3-1}{\sqrt3+1}\to\boxed{\bold{D.}}[/tex]
