Question

Evaluate the expression:
$\frac{1}{(3-\sqrt8)}-\frac{1}{(\sqrt8-\sqrt7)}+\frac{1}{(\sqrt7-\sqrt6)}-\frac{1}{(\sqrt6-\sqrt5)}+\frac{1}{(\sqrt5-2)}$

• A. 5
• B. 3
• C. 2
• D. 0

Answer 1

Answer:

A. 5

Rationalize each term using:

$\frac{1}{a-b}=\frac{a+b}{a^2-b^2}$

Now simplify term by term:

$\frac{1}{3-\sqrt8}$
$= \frac{3+\sqrt8}{9-8}$
$= 3+\sqrt8$

$\frac{1}{\sqrt8-\sqrt7}$
$= \frac{\sqrt8+\sqrt7}{8-7}$
$= \sqrt8+\sqrt7$
$\frac{1}{\sqrt7-\sqrt6}$
$= \sqrt7+\sqrt6$

$\frac{1}{\sqrt6-\sqrt5}$
$= \sqrt6+\sqrt5$

$\frac{1}{\sqrt5-2}$
$= \frac{\sqrt5+2}{5-4}$
$= \sqrt5+2$

Substitute back:

$(3+\sqrt8)-(\sqrt8+\sqrt7)+(\sqrt7+\sqrt6)-(\sqrt6+\sqrt5)+(\sqrt5+2)$

Now cancel common terms:

$3+\cancel{\sqrt8}-\cancel{\sqrt8}-\cancel{\sqrt7}$
$+\cancel{\sqrt7}+\cancel{\sqrt6}$
$-\cancel{\sqrt6}-\cancel{\sqrt5}$
$+\cancel{\sqrt5}+2$
]Remaining:

$3+2=5$

$\boxed{5}$