Answer
$\dfrac{\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}}=3-2\sqrt{2}+2\sqrt{3}-\sqrt{6}$
Work Step by Step
$\dfrac{\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}}$
Multiply the numerator and the denominator of this expression by the conjugate of the denominator and simplify if possible:
$\dfrac{\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}}=\dfrac{\sqrt{3}+\sqrt{4}}{\sqrt{2}+\sqrt{3}}\cdot\dfrac{\sqrt{2}-\sqrt{3}}{\sqrt{2}-\sqrt{3}}=\dfrac{(\sqrt{3}+\sqrt{4})(\sqrt{2}-\sqrt{3})}{(\sqrt{2})^{2}-(\sqrt{3})^{2}}=...$
$...=\dfrac{\sqrt{6}-\sqrt{3^{2}}+\sqrt{8}-\sqrt{12}}{2-3}=\dfrac{\sqrt{6}-3+2\sqrt{2}-2\sqrt{3}}{-1}=...$
$...=3-2\sqrt{2}+2\sqrt{3}-\sqrt{6}$
