1.

3+√6/√3+√2=a+√3b find the value of b​

Answer»

ANSWER:

3 + \frac{ \sqrt{6} }{ \sqrt{3} + \sqrt{<klux>2</klux>} } \\ = 3 + \frac{ \sqrt{6}( \sqrt{3} - \sqrt{2} ) }{( \sqrt{3} + \sqrt{2})( \sqrt{3} - \sqrt{2} ) } \\ = 3 + \frac{3 \sqrt{2} - 2 \sqrt{3} }{3 - 2} \\ = 3 + \sqrt{3} ( \sqrt{6} - 2) = a + \sqrt{3}b

THEREFORE,

a = 3

b = √6 - 2


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