If frac{4 sqrt{3}+5 sqrt{2}}{sqrt{48}+sqrt{18 | vedclass

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(D) Given expression: $\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$Simplify the denominator: $\sqrt{48} = \sqrt{16 	imes 3} = 4 \sqrt{3}$ and $

Vedclass · Accepted Answer

$\frac{3}{5}, \frac{4}{15}$

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(D) Given expression: $\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}$Simplify the denominator: $\sqrt{48} = \sqrt{16 	imes 3} = 4 \sqrt{3}$ and $\sqrt{18} = \sqrt{9 	imes 2} = 3 \sqrt{2}$.So,the expression becomes: $\frac{4 \sqrt{3}+5 \sqrt{2}}{4 \sqrt{3}+3 \sqrt{2}}$.Rationalize the denominator by multiplying the numerator and denominator by $(4 \sqrt{3}-3 \sqrt{2})$:$= \frac{(4 \sqrt{3}+5 \sqrt{2})(4 \sqrt{3}-3 \sqrt{2})}{(4 \sqrt{3}+3 \sqrt{2})(4 \sqrt{3}-3 \sqrt{2})}$Expand the numerator: $(4 \sqrt{3})(4 \sqrt{3}) - (4 \sqrt{3})(3 \sqrt{2}) + (5 \sqrt{2})(4 \sqrt{3}) - (5 \sqrt{2})(3 \sqrt{2})$$= 16(3) - 12 \sqrt{6} + 20 \sqrt{6} - 15(2) = 48 + 8 \sqrt{6} - 30 = 18 + 8 \sqrt{6}$.Expand the denominator: $(4 \sqrt{3})^2 - (3 \sqrt{2})^2 = 16(3) - 9(2) = 48 - 18 = 30$.Thus,the expression is: $\frac{18 + 8 \sqrt{6}}{30} = \frac{18}{30} + \frac{8 \sqrt{6}}{30} = \frac{3}{5} + \frac{4}{15} \sqrt{6}$.Comparing this with $a+b \sqrt{6}$,we get $a = \frac{3}{5}$ and $b = \frac{4}{15}$.

If $\frac{4 \sqrt{3}+5 \sqrt{2}}{\sqrt{48}+\sqrt{18}}=a+b \sqrt{6},$ then the values of $a$ and $b$ are,respectively

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