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After rationalizing the denominator of \frac{7}{3\sqrt{3}-2\sqrt{2}} we get the denominator as:


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Answer.    [B]
We have,\frac{7}{3\sqrt{3}-2\sqrt{2}}
We have to rationalize it

\frac{7}{3\sqrt{3}-2\sqrt{2}}\times \frac{3\sqrt{3}+2\sqrt{2}}{3\sqrt{3}+2\sqrt{2}}       
[Multiplying numerator and denominator by 3\sqrt{3}+2\sqrt{2}]

= \frac{7\left ( 3\sqrt{3}+2\sqrt{2} \right )}{\left ( 3\sqrt{3} \right )^{2}-\left ( 2\sqrt{2} \right )^{2}}                    [\because (a – b) (a + b) = a2 – b2]

\frac{7\left ( 3\sqrt{3} \right )+2\sqrt{2}}{27-8}

\frac{7\left ( 3\sqrt{3} \right )+2\sqrt{2}}{19}
Therefore we get the denominator as 19.
Hence (B) is the correct option.

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