Rationalise the denominator of √3 + √2/5 + √2

1.	Rationalise the denominator of √3 + √2/5 + √2
Answer» Step-by-step EXPLANATION: hey here is your SOLUTION pls mark it as BRAINLIEST $so \: here \: \\ √3 + √2/5 + √2 \\ so \: conjugate \: of \: 5 + \sqrt{2} \: \: is \: 5 - \sqrt{2} \\ so \: rationalising \: the \: denominator \: ie \: by \: conjugate \: method \\ we \: get \\ ( \sqrt{3} + \sqrt{2} ) \: \: \times ( 5 - \sqrt{2} ) \div 5 + \sqrt{2} \times 5 - \sqrt{2} \\ = 5 \sqrt{3} - \sqrt{6} + 5 \sqrt{2} - 2 \div 25 - 2 \\ = 5 \sqrt{3} - \sqrt{6} + 5 \sqrt{2} - 2 \div 23$

Answer»

Step-by-step EXPLANATION:

hey here is your SOLUTION

pls mark it as BRAINLIEST

$so \: here \: \\ √3 + √2/5 + √2 \\ so \: conjugate \: of \: 5 + \sqrt{2} \: \: is \: 5 - \sqrt{2} \\ so \: rationalising \: the \: denominator \: ie \: by \: conjugate \: method \\ we \: get \\ ( \sqrt{3} + \sqrt{2} ) \: \: \times ( 5 - \sqrt{2} ) \div 5 + \sqrt{2} \times 5 - \sqrt{2} \\ = 5 \sqrt{3} - \sqrt{6} + 5 \sqrt{2} - 2 \div 25 - 2 \\ = 5 \sqrt{3} - \sqrt{6} + 5 \sqrt{2} - 2 \div 23$

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Rationalise the denominator of √3 + √2/5 + √2

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