frac{sqrt{24}+sqrt{216}}{sqrt{96}}=? | vedclass

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(C) To solve the expression $\frac{\sqrt{24}+\sqrt{216}}{\sqrt{96}}$,we simplify each radical term by factoring out the perfect squares:$\sqrt{24} = \

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$2$

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(C) To solve the expression $\frac{\sqrt{24}+\sqrt{216}}{\sqrt{96}}$,we simplify each radical term by factoring out the perfect squares:$\sqrt{24} = \sqrt{4 	imes 6} = 2\sqrt{6}$$\sqrt{216} = \sqrt{36 	imes 6} = 6\sqrt{6}$$\sqrt{96} = \sqrt{16 	imes 6} = 4\sqrt{6}$Now,substitute these values back into the expression:$\frac{2\sqrt{6} + 6\sqrt{6}}{4\sqrt{6}} = \frac{(2+6)\sqrt{6}}{4\sqrt{6}} = \frac{8\sqrt{6}}{4\sqrt{6}}$Canceling $\sqrt{6}$ from the numerator and denominator:$\frac{8}{4} = 2$

$\frac{\sqrt{24}+\sqrt{216}}{\sqrt{96}}=?$

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