Answer
$\dfrac{2\sqrt{3}+\sqrt{6}}{4\sqrt{3}-\sqrt{6}}=\dfrac{3\sqrt{2}+5}{7}$
Work Step by Step
$\dfrac{2\sqrt{3}+\sqrt{6}}{4\sqrt{3}-\sqrt{6}}$
Multiply the numerator and the denominator of this expression by the conjugate of the denominator:
$\dfrac{2\sqrt{3}+\sqrt{6}}{4\sqrt{3}-\sqrt{6}}=\dfrac{2\sqrt{3}+\sqrt{6}}{4\sqrt{3}-\sqrt{6}}\cdot\dfrac{4\sqrt{3}+\sqrt{6}}{4\sqrt{3}+\sqrt{6}}=...$
$...=\dfrac{(2\sqrt{3}+\sqrt{6})(4\sqrt{3}+\sqrt{6})}{(4\sqrt{3})^{2}-(\sqrt{6})^{2}}=...$
$...=\dfrac{8\sqrt{3^{2}}+2\sqrt{18}+4\sqrt{18}+\sqrt{6^{2}}}{16(3)-6}=...$
$...=\dfrac{8(3)+6\sqrt{18}+6}{48-6}=\dfrac{6(3)\sqrt{2}+24+6}{42}=\dfrac{18\sqrt{2}+30}{42}=...$
Take out common factor $6$ from the numerator and simplify:
$...=\dfrac{6(3\sqrt{2}+5)}{42}=\dfrac{3\sqrt{2}+5}{7}$
