Answer:
[tex]\large\boxed{\dfrac{\sqrt2+\sqrt5}{\sqrt2-\sqrt5}=-\dfrac{7+2\sqrt{10}}{3}}[/tex]
Step-by-step explanation:
[tex]\text{Use}\ (a-b)(a+b)=a^2-b^2\ \text{and}\ (a+b)^2=a^2+2ab+b^2\\\\\\\dfrac{\sqrt2+\sqrt5}{\sqrt2-\sqrt5}=\dfrac{\sqrt2+\sqrt5}{\sqrt2-\sqrt5}\cdot\dfrac{\sqrt2+\sqrt5}{\sqrt2+\sqrt5}=\dfrac{(\sqrt2+\sqrt5)^2}{(\sqrt2)^2-(\sqrt5)^2}\\\\=\dfrac{(\sqrt2)^2+2(\sqrt2)(\sqrt5)+(\sqrt5)^2}{2-5}=\dfrac{2+2\sqrt{(2)(5)}+5}{-3}\\\\=-\dfrac{7+2\sqrt{10}}{3}[/tex]
