If sqrt{left(1+frac{27}{169}right)}=1+frac{x} | vedclass

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(A) Given the equation: $\sqrt{\left(1+\frac{27}{169}\right)}=1+\frac{x}{13}$First,simplify the expression inside the square root:$1+\frac{27}{169} =

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(A) Given the equation: $\sqrt{\left(1+\frac{27}{169}\right)}=1+\frac{x}{13}$First,simplify the expression inside the square root:$1+\frac{27}{169} = \frac{169+27}{169} = \frac{196}{169}$Now,take the square root of the fraction:$\sqrt{\frac{196}{169}} = \frac{14}{13}$Substitute this back into the original equation:$\frac{14}{13} = 1+\frac{x}{13}$Subtract $1$ from both sides:$\frac{14}{13} - 1 = \frac{x}{13}$$\frac{14-13}{13} = \frac{x}{13}$$\frac{1}{13} = \frac{x}{13}$Therefore,$x = 1$.

If $\sqrt{\left(1+\frac{27}{169}\right)}=1+\frac{x}{13},$ then $x$ equals

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