Question

Rationalize the denominator
3√2/2√3+√5​

Answer 1

Answer:

$\frac{6\sqrt{6} - 3\sqrt{10} }{7}$ (denominator has been rationalised )

Step-by-step explanation:

$\frac{3\sqrt{2} }{2\sqrt{3}+\sqrt{5} }$

= $\frac{3\sqrt{2} }{2\sqrt{3}+\sqrt{5} }$ * $\frac{2\sqrt{3}-\sqrt{5} }{2\sqrt{3}-\sqrt{5} }$

= $\frac{3\sqrt{2}*(2\sqrt{3}-\sqrt{5}) }{(2\sqrt{3}+\sqrt{5})(2\sqrt{3}-\sqrt{5} ) }$

= $\frac{(3\sqrt{2}*2\sqrt{3}) - (3\sqrt{2}*\sqrt{5}) }{(2\sqrt{3})^2-(\sqrt{5})^2 }$

= $\frac{6\sqrt{6} - 3\sqrt{10} }{(4*3) - 5 }$

= $\frac{6\sqrt{6} - 3\sqrt{10} }{12- 5 }$

= $\frac{6\sqrt{6} - 3\sqrt{10} }{7}$  , here the denominator has been rationalised