[tex] \dfrac{3}{2 + \sqrt{2} } [/tex]
To rationalise the denominator, this is a standard step:
[tex] = \dfrac{3}{2 + \sqrt{2} } \times \dfrac{2 - \sqrt{2}}{2 - \sqrt{2}} [/tex]
Answer: Option A
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Let me continue to complete the question, though it is not ask for
[tex]= \dfrac{3(2 - \sqrt{2})}{(2 + \sqrt{2)}(2 - \sqrt{2}) } [/tex]
[tex]= \dfrac{3(2 - \sqrt{2})}{4 - 2 } [/tex]
[tex]= \dfrac{6 - 2\sqrt{2}}{2 } [/tex]
