If sqrt{1+sqrt{1-frac{2176}{2401}}}=1+frac{x} | vedclass

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Question

(B) Given the equation: $\sqrt{1+\sqrt{1-\frac{2176}{2401}}}=1+\frac{x}{7}$First,simplify the term inside the inner square root: $1-\frac{2176}{2401}

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$1$

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(B) Given the equation: $\sqrt{1+\sqrt{1-\frac{2176}{2401}}}=1+\frac{x}{7}$First,simplify the term inside the inner square root: $1-\frac{2176}{2401} = \frac{2401-2176}{2401} = \frac{225}{2401}$Now,take the square root of this fraction: $\sqrt{\frac{225}{2401}} = \frac{15}{49}$Substitute this back into the original equation: $\sqrt{1+\frac{15}{49}} = 1+\frac{x}{7}$Simplify the expression inside the outer square root: $\sqrt{\frac{49+15}{49}} = \sqrt{\frac{64}{49}}$Calculate the square root: $\frac{8}{7} = 1+\frac{x}{7}$Rewrite $\frac{8}{7}$ as $1+\frac{1}{7}$: $1+\frac{1}{7} = 1+\frac{x}{7}$Comparing both sides,we get $x = 1$.

If $\sqrt{1+\sqrt{1-\frac{2176}{2401}}}=1+\frac{x}{7},$ the value of $x$ is

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