Find the value of a in the following:frac{6}{ | vedclass

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(B) To solve $\frac{6}{3 \sqrt{2}-2 \sqrt{3}}$,we rationalize the denominator by multiplying the numerator and the denominator by the conjugate $(3 \s

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$-2$

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(B) To solve $\frac{6}{3 \sqrt{2}-2 \sqrt{3}}$,we rationalize the denominator by multiplying the numerator and the denominator by the conjugate $(3 \sqrt{2}+2 \sqrt{3})$.$\frac{6}{3 \sqrt{2}-2 \sqrt{3}} = \frac{6(3 \sqrt{2}+2 \sqrt{3})}{(3 \sqrt{2}-2 \sqrt{3})(3 \sqrt{2}+2 \sqrt{3})}$Using the identity $(x-y)(x+y) = x^2 - y^2$,the denominator becomes $(3 \sqrt{2})^2 - (2 \sqrt{3})^2 = (9 	imes 2) - (4 	imes 3) = 18 - 12 = 6$.So,the expression simplifies to $\frac{6(3 \sqrt{2}+2 \sqrt{3})}{6} = 3 \sqrt{2} + 2 \sqrt{3}$.Comparing this with the given equation $3 \sqrt{2} - a \sqrt{3} = 3 \sqrt{2} + 2 \sqrt{3}$,we get $-a \sqrt{3} = 2 \sqrt{3}$.Dividing both sides by $-\sqrt{3}$,we find $a = -2$.

Find the value of $a$ in the following:
$\frac{6}{3 \sqrt{2}-2 \sqrt{3}}=3 \sqrt{2}-a \sqrt{3}$

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Find the value of $a$ in the following:$\frac{6}{3 \sqrt{2}-2 \sqrt{3}}=3 \sqrt{2}-a \sqrt{3}$