1.

4root3+5root2/root48+root18

Answer»

ANSWER:

To rationalise the denominator of

\frac{4 \sqrt{3} + 5 \sqrt{2} }{ \sqrt{48} + \sqrt{18} }

It can be written as

\frac{4 \sqrt{3} + 5 \sqrt{2} }{4 \sqrt{3} - 3 \sqrt{2} } \\ \\ \frac{4 \sqrt{3} + 5 \sqrt{2} }{4 \sqrt{3} - 3 \sqrt{2} } \times \frac{4 \sqrt{3} + 3 \sqrt{2} }{4 \sqrt{3} + 3 \sqrt{2} } \\ \\ \frac{4 \sqrt{3} (4 \sqrt{3} + 3 \sqrt{2}) + 5 \sqrt{2} (4 \sqrt{3} + 3 \sqrt{2}) }{(4 \sqrt{3} ) {}^{2} - (3 \sqrt{2}) {}^{2} } \\ \\ \frac{48 + 12 \sqrt{6} + 20 \sqrt{6} + 30}{48 - 18} \\ \\ \frac{78 + 32 \sqrt{6} }{30} \\ \\ \frac{ \cancel{2 \: }\:(39 + 16 \sqrt{6} )}{ \cancel{30} \: \: {}^{15} } \\ \\ \boxed{ \BOLD{ \frac{39 + 16 \sqrt{6} }{15} }}

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