Question

Rationalise The denominator of 3+√2\2(√3+√2)​

Answer 1

$\frac{3 + \sqrt{2} }{2 \sqrt{3} + 2 \sqrt{2} } \\ ( \frac{3 + \sqrt{2} }{2 \sqrt{3} + 2 \sqrt{2} } ) \times ( \frac{2 \sqrt{3 } - 2 \sqrt{2} }{2 \sqrt{3} - 2 \sqrt{2} } ) \\ \frac{(3 + \sqrt{2} )(2 \sqrt{3} - 2 \sqrt{2}) }{(2 \sqrt{3}) ^{2} - (2 \sqrt{2} ^{2}) } \\ \frac{6 \sqrt{3} - 3\sqrt{2} + 2 \sqrt{6} - 2 \times 2 }{4 \times 3 - 4 \times 2} \\ \frac{6 \sqrt{3} - 3 \sqrt{2} + 2 \sqrt{6} - 4}{4}$

Answer 2

To rationalize the denominator of (3 + √2)/2(√3 + √2) we will multiply (√3 - √2) with both numerator and denominator.

So we will get (3 + √2)(√3 - √2)/2(√3 + √2)(√3 - √2)

= [3√3 - 3√2 + √6 - 2] / 2[9 - 4]

= [3√3 - 3√2 + √6 - 2] / 10

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