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Rationalize the denominator of [tex]\frac{\sqrt{6}}{\sqrt{2}+\sqrt{3}}[/tex]

Sagot :

To rationalize the denominator of the expression [tex]\(\frac{\sqrt{6}}{\sqrt{2} + \sqrt{3}}\)[/tex], we need to eliminate the surds (square roots) in the denominator. Here's the step-by-step process:

### Step 1: Multiply by the Conjugate
The conjugate of [tex]\(\sqrt{2} + \sqrt{3}\)[/tex] is [tex]\(\sqrt{2} - \sqrt{3}\)[/tex]. Multiplying the numerator and the denominator by [tex]\(\sqrt{2} - \sqrt{3}\)[/tex] will help us rationalize the denominator.

[tex]\[ \frac{\sqrt{6}}{\sqrt{2} + \sqrt{3}} \times \frac{\sqrt{2} - \sqrt{3}}{\sqrt{2} - \sqrt{3}} = \frac{\sqrt{6} (\sqrt{2} - \sqrt{3})}{(\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3})} \][/tex]

### Step 2: Simplify the Denominator
Use the difference of squares formula [tex]\((a + b)(a - b) = a^2 - b^2\)[/tex]:

[tex]\[ (\sqrt{2} + \sqrt{3})(\sqrt{2} - \sqrt{3}) = (\sqrt{2})^2 - (\sqrt{3})^2 = 2 - 3 = -1 \][/tex]

### Step 3: Simplify the Numerator
Distribute [tex]\(\sqrt{6}\)[/tex] across [tex]\(\sqrt{2} - \sqrt{3}\)[/tex]:

[tex]\[ \sqrt{6} (\sqrt{2} - \sqrt{3}) = \sqrt{6 \times 2} - \sqrt{6 \times 3} = \sqrt{12} - \sqrt{18} \][/tex]

We can simplify [tex]\(\sqrt{12}\)[/tex] and [tex]\(\sqrt{18}\)[/tex]:

[tex]\[ \sqrt{12} = \sqrt{4 \times 3} = 2\sqrt{3} \][/tex]
[tex]\[ \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2} \][/tex]

So, the numerator becomes:

[tex]\[ 2\sqrt{3} - 3\sqrt{2} \][/tex]

### Step 4: Combine and Simplify the Entire Expression
Putting it all together:

[tex]\[ \frac{2\sqrt{3} - 3\sqrt{2}}{-1} \][/tex]

Dividing each term in the numerator by [tex]\(-1\)[/tex]:

[tex]\[ \frac{2\sqrt{3}}{-1} - \frac{3\sqrt{2}}{-1} = -2\sqrt{3} + 3\sqrt{2} \][/tex]

Thus, the rationalized form of the given expression is:

[tex]\[ \frac{\sqrt{6}}{\sqrt{2} + \sqrt{3}} = 3\sqrt{2} - 2\sqrt{3} \][/tex]

This is the final simplified result with a rationalized denominator.
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