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(A) To find the value of $\frac{1}{\sqrt{3} + \sqrt{2}}$,we first rationalize the denominator by multiplying the numerator and the denominator by the

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$0.318$

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(A) To find the value of $\frac{1}{\sqrt{3} + \sqrt{2}}$,we first rationalize the denominator by multiplying the numerator and the denominator by the conjugate $(\sqrt{3} - \sqrt{2})$.$\frac{1}{\sqrt{3} + \sqrt{2}} 	imes \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} - \sqrt{2}} = \frac{\sqrt{3} - \sqrt{2}}{(\sqrt{3})^2 - (\sqrt{2})^2}$$= \frac{\sqrt{3} - \sqrt{2}}{3 - 2} = \frac{\sqrt{3} - \sqrt{2}}{1} = \sqrt{3} - \sqrt{2}$Given $\sqrt{3} = 1.732$ and $\sqrt{2} = 1.414$,we substitute these values:$1.732 - 1.414 = 0.318$Thus,the value is $0.318$.

Find the value of the following expression correct to three decimal places,rationalizing the denominator if needed. Given $\sqrt{2} = 1.414$,$\sqrt{3} = 1.732$,and $\sqrt{5} = 2.236$.
$\frac{1}{\sqrt{3} + \sqrt{2}}$

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Find the value of the following expression correct to three decimal places,rationalizing the denominator if needed. Given $\sqrt{2} = 1.414$,$\sqrt{3} = 1.732$,and $\sqrt{5} = 2.236$.$\frac{1}{\sqrt{3} + \sqrt{2}}$