Answer :
Answer:
[tex]\huge\boxed{x+\dfrac{1}{x}=(13+2\sqrt{42})+(13-2\sqrt{42})=26}[/tex]
Step-by-step explanation:
[tex]x=\dfrac{\sqrt7+\sqrt6}{\sqrt7-\sqrt6}\cdot\dfrac{\sqrt7+\sqrt6}{\sqrt7+\sqrt6}=\dfrac{(\sqrt7+\sqrt6)^2}{(\sqrt7-\sqrt6)(\sqrt6+\sqrt7)}\\\\\\\text{use}\ (a+b)^2=a^2+2ab+b^2\ \text{and}\ (a-b)(a+b)=a^2-b^2[/tex]
[tex]x=\dfrac{(\sqrt7)^2+(2)(\sqrt7)(\sqrt6)+(\sqrt6)^2}{(\sqrt7)^2-(\sqrt6)^2}\\\\\\\text{use}\ (\sqrt{a})^2=a,\ \text{for}\ a\geq0\\\\\\x=\dfrac{7+2\sqrt{42}+6}{7-6}=\dfrac{13+2\sqrt{42}}{1}=13+2\sqrt{42}\\\\\dfrac{1}{x}=\dfrac{1}{13+2\sqrt{42}}\cdot\dfrac{13-2\sqrt{42}}{13-2\sqrt{42}}=\dfrac{13-2\sqrt{42}}{13^2-(2\sqrt{42})^2}=\dfrac{13-2\sqrt{42}}{169-168}\\\\\dfrac{1}{x}=13-2\sqrt{42}[/tex]