`1/(sqrt3+sqrt4)+1/(sqrt4+sqrt5)+1/(sqrt5+sqrt6)+1/(sqrt6+sqrt7)+1/(sqrt7+sqrt8)+1/(sqrt8+sqrt9)` is equal to?

Options

A. a)`sqrt3`
B. b)`3sqrt3`
C. c)`3-sqrt3`
D. d)`5-sqrt3`

Answer

To solve the expression

\[ \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}}, \]

we will rationalize each term and simplify the entire expression step by step.

### Step 1: Rationalize the first term
The first term is

\[ \frac{1}{\sqrt{3} + \sqrt{4}}. \]

To rationalize it, we multiply the numerator and denominator by the conjugate of the denominator, which is \(\sqrt{3} - \sqrt{4}\):

\[ \frac{1}{\sqrt{3} + \sqrt{4}} \cdot \frac{\sqrt{3} - \sqrt{4}}{\sqrt{3} - \sqrt{4}} = \frac{\sqrt{3} - \sqrt{4}}{(\sqrt{3})^2 - (\sqrt{4})^2} = \frac{\sqrt{3} - \sqrt{4}}{3 - 4} = \frac{\sqrt{3} - \sqrt{4}}{-1} = \sqrt{4} - \sqrt{3} = 2 - \sqrt{3}. \]

### Step 2: Rationalize the second term
The second term is

\[ \frac{1}{\sqrt{4} + \sqrt{5}}. \]

Rationalizing gives:

\[ \frac{1}{\sqrt{4} + \sqrt{5}} \cdot \frac{\sqrt{4} - \sqrt{5}}{\sqrt{4} - \sqrt{5}} = \frac{\sqrt{4} - \sqrt{5}}{(\sqrt{4})^2 - (\sqrt{5})^2} = \frac{\sqrt{4} - \sqrt{5}}{4 - 5} = \frac{\sqrt{4} - \sqrt{5}}{-1} = \sqrt{5} - \sqrt{4} = \sqrt{5} - 2. \]

### Step 3: Rationalize the third term
The third term is

\[ \frac{1}{\sqrt{5} + \sqrt{6}}. \]

Rationalizing gives:

\[ \frac{1}{\sqrt{5} + \sqrt{6}} \cdot \frac{\sqrt{5} - \sqrt{6}}{\sqrt{5} - \sqrt{6}} = \frac{\sqrt{5} - \sqrt{6}}{(\sqrt{5})^2 - (\sqrt{6})^2} = \frac{\sqrt{5} - \sqrt{6}}{5 - 6} = \frac{\sqrt{5} - \sqrt{6}}{-1} = \sqrt{6} - \sqrt{5}. \]

### Step 4: Rationalize the fourth term
The fourth term is

\[ \frac{1}{\sqrt{6} + \sqrt{7}}. \]

Rationalizing gives:

\[ \frac{1}{\sqrt{6} + \sqrt{7}} \cdot \frac{\sqrt{6} - \sqrt{7}}{\sqrt{6} - \sqrt{7}} = \frac{\sqrt{6} - \sqrt{7}}{(\sqrt{6})^2 - (\sqrt{7})^2} = \frac{\sqrt{6} - \sqrt{7}}{6 - 7} = \frac{\sqrt{6} - \sqrt{7}}{-1} = \sqrt{7} - \sqrt{6}. \]

### Step 5: Rationalize the fifth term
The fifth term is

\[ \frac{1}{\sqrt{7} + \sqrt{8}}. \]

Rationalizing gives:

\[ \frac{1}{\sqrt{7} + \sqrt{8}} \cdot \frac{\sqrt{7} - \sqrt{8}}{\sqrt{7} - \sqrt{8}} = \frac{\sqrt{7} - \sqrt{8}}{(\sqrt{7})^2 - (\sqrt{8})^2} = \frac{\sqrt{7} - \sqrt{8}}{7 - 8} = \frac{\sqrt{7} - \sqrt{8}}{-1} = \sqrt{8} - \sqrt{7}. \]

### Step 6: Rationalize the sixth term
The sixth term is

\[ \frac{1}{\sqrt{8} + \sqrt{9}}. \]

Rationalizing gives:

\[ \frac{1}{\sqrt{8} + \sqrt{9}} \cdot \frac{\sqrt{8} - \sqrt{9}}{\sqrt{8} - \sqrt{9}} = \frac{\sqrt{8} - \sqrt{9}}{(\sqrt{8})^2 - (\sqrt{9})^2} = \frac{\sqrt{8} - \sqrt{9}}{8 - 9} = \frac{\sqrt{8} - \sqrt{9}}{-1} = \sqrt{9} - \sqrt{8} = 3 - \sqrt{8}. \]

### Step 7: Combine all terms
Now we combine all the results:

\[ (2 - \sqrt{3}) + (\sqrt{5} - 2) + (\sqrt{6} - \sqrt{5}) + (\sqrt{7} - \sqrt{6}) + (\sqrt{8} - \sqrt{7}) + (3 - \sqrt{8}). \]

Notice that all intermediate terms cancel out:

- \(-2\) and \(+2\) cancel.
- \(-\sqrt{5}\) and \(+\sqrt{5}\) cancel.
- \(-\sqrt{6}\) and \(+\sqrt{6}\) cancel.
- \(-\sqrt{7}\) and \(+\sqrt{7}\) cancel.
- \(-\sqrt{8}\) and \(+\sqrt{8}\) cancel.

What remains is:

\[ 3 - \sqrt{3}. \]

### Final Answer
Thus, the value of the entire expression is

\[ 3 - \sqrt{3}. \]