1.77 If $x=\sqrt{7}+\sqrt{6}$, then $\left(x+\frac{1}{x}\right):\left(x-\frac{1}{x}\right)$ equals
(A) $\sqrt{7}: \sqrt{6}$.
(B) $7: 6$.
(C) $x^{2}: 1$.
(D) $x$.
(1st "Five Sheep Cup" Junior High School Mathematics Competition, 1989)
［Solution］Given that
$$
\begin{aligned}
\frac{1}{x} & =\frac{1}{\sqrt{7}+\sqrt{6}}=\frac{\sqrt{7}-\sqrt{6}}{(\sqrt{7})^{2}-(\sqrt{6})^{2}} \\
& =\sqrt{7}-\sqrt{6} .
\end{aligned}
$$

Then $\left(x+\frac{1}{x}\right):\left(x-\frac{1}{x}\right)=2 \sqrt{7}: 2 \sqrt{6}=\sqrt{7}: \sqrt{6}$. Therefore, the answer is $(A)$.