If $$\frac{7 + 2\sqrt{5}}{11 + 3\sqrt{5}}$$  = a + b$$\sqrt{5}$$, then:​

Correct option is D

Given:

​$$\frac{7 + 2\sqrt{5}}{11 + 3\sqrt{5}}$$ = a + b$$\sqrt{5}$$​

Solution:

To rationalize the denominator, we multiply by the conjugate of the denominator.

$$\begin{aligned}& \frac{(7+2 \sqrt{5})(11-3 \sqrt{5})}{(11+3 \sqrt{5})(11-3 \sqrt{5})} \\& \Rightarrow \frac{7 \times 11+7 \times(-3 \sqrt{5})+2 \sqrt{5} \times 11+2 \sqrt{5} \times(-3 \sqrt{5})}{11^2-(3 \sqrt{5})^2} \\& \Rightarrow \frac{77-21 \sqrt{5}+22 \sqrt{5}-30}{121-45} \\& \Rightarrow \frac{47+\sqrt{5}}{76} \\& \Rightarrow \frac{47}{76}+\frac{1}{76} \sqrt{5}\end{aligned}$$    

Comparing with a + b$$\sqrt{5}$$ 

a = $$\frac{47}{76}$$ , b =$$\frac{1}{76}$$​​

Thus, the correct answer is (D)

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