Question

Simplify
$(\frac{1}{7+\sqrt2})+(\frac{1}{7-\sqrt2})-(\frac{14}{49-2})$

• A. 0
• B. 1
• C. 2
• D. √2

Answer 1

Answer:

A. 0

Simplify:

$\left(\frac{1}{7+\sqrt{2}}\right)+\left(\frac{1}{7-\sqrt{2}}\right)-\left(\frac{14}{49-2}\right)$

First, combine the first two fractions:

$\frac{1}{7+\sqrt2}+\frac{1}{7-\sqrt2}$

Using:

$\frac{1}{a+b}+\frac{1}{a-b}$

$\frac{(a-b)+(a+b)}{a^2-b^2}$

$\frac{2a}{a^2-b^2}$

Here $(a=7,\ b=\sqrt2)$:

$= \frac{14}{49-2}$

$= \frac{14}{47}$

Now subtract the third term:

$\frac{14}{47}-\frac{14}{47}=0$

Therefore,

$\boxed{0}$