frac{sqrt{7}}{sqrt{16+6 sqrt{7}}-sqrt{16-6 sq | vedclass

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Question

(A) First,simplify the terms inside the square roots in the denominator.$\sqrt{16+6 \sqrt{7}} = \sqrt{9+7+2 	imes 3 	imes \sqrt{7}} = \sqrt{3^2 + (\

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$\frac{1}{2}$

Vedclass · Answer

(A) First,simplify the terms inside the square roots in the denominator.$\sqrt{16+6 \sqrt{7}} = \sqrt{9+7+2 	imes 3 	imes \sqrt{7}} = \sqrt{3^2 + (\sqrt{7})^2 + 2 	imes 3 	imes \sqrt{7}} = \sqrt{(3+\sqrt{7})^2} = 3+\sqrt{7}$.Similarly,$\sqrt{16-6 \sqrt{7}} = \sqrt{9+7-2 	imes 3 	imes \sqrt{7}} = \sqrt{(3-\sqrt{7})^2} = 3-\sqrt{7}$.Now,substitute these values into the denominator:$\sqrt{16+6 \sqrt{7}} - \sqrt{16-6 \sqrt{7}} = (3+\sqrt{7}) - (3-\sqrt{7}) = 3 + \sqrt{7} - 3 + \sqrt{7} = 2\sqrt{7}$.Finally,substitute this back into the original expression:$\frac{\sqrt{7}}{2\sqrt{7}} = \frac{1}{2}$.

$\frac{\sqrt{7}}{\sqrt{16+6 \sqrt{7}}-\sqrt{16-6 \sqrt{7}}}$ is equal to

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