If k > 1, which of the following must be equal to
$$\frac{2}{\sqrt{k+1}+\sqrt{k-1}}?$$
$$A.\ 2$$
$$B.\ 2\sqrt{k}$$
$$C.\ 2\sqrt{k+1}+\sqrt{k-1}$$
$$D.\ \frac{\sqrt{k+1}}{\sqrt{k-1}}$$
$$E.\ \sqrt{k+1}-\sqrt{k-1}$$
The OA is E.
I know that I can solve this PS question in an easy way like,
$$\frac{2}{\sqrt{k+1}+\sqrt{k-1}}\cdot\frac{\sqrt{k+1}-\sqrt{k-1}}{\sqrt{k+1}-\sqrt{k-1}}=\frac{2\sqrt{k+1}-\sqrt{k-1}}{2}=\sqrt{k+1}-\sqrt{k-1}$$
Please, can any expert explain this PS question for me? I would like to know another way to solve it. I need your help. Thanks.
$$\frac{2}{\sqrt{k+1}+\sqrt{k-1}}?$$
$$A.\ 2$$
$$B.\ 2\sqrt{k}$$
$$C.\ 2\sqrt{k+1}+\sqrt{k-1}$$
$$D.\ \frac{\sqrt{k+1}}{\sqrt{k-1}}$$
$$E.\ \sqrt{k+1}-\sqrt{k-1}$$
The OA is E.
I know that I can solve this PS question in an easy way like,
$$\frac{2}{\sqrt{k+1}+\sqrt{k-1}}\cdot\frac{\sqrt{k+1}-\sqrt{k-1}}{\sqrt{k+1}-\sqrt{k-1}}=\frac{2\sqrt{k+1}-\sqrt{k-1}}{2}=\sqrt{k+1}-\sqrt{k-1}$$
Please, can any expert explain this PS question for me? I would like to know another way to solve it. I need your help. Thanks.
















