Question

If
$\sqrt{1 + \frac{27}{169 } } = 1 + \frac{x}{13}$
find the value of x.




plz give me answer I will mark as BRAIN LIST.

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Answer 1

Answer:

$=> \sqrt{1 + \frac{27}{169} } = 1 + \frac{x}{13} \\ => \sqrt{ \frac{196}{169} } = \frac{13 + x}{13} \\ = > \frac{14}{13} = \frac{x + 13}{13} \\ = > 14 = x + 13 \\ = > x = 1$

Answer 2

Given,

$\sf{\sqrt{1+\dfrac{27}{169} }=1+\dfrac{x}{13} }$

To Find :-

The value of x.

Solution :-

$\sf{\sqrt{1+\dfrac{27}{169} }=1+\dfrac{x}{13} }\\\\\displaystyle{\sf{\longrightarrow \sqrt{\frac{169+27}{169} } =1+\frac{x}{13} }}\\\\\displaystyle{\sf{\longrightarrow \sqrt{\frac{196}{169}}=\frac{13+x}{13} }}\\\\\displaystyle{\sf{\longrightarrow \frac{14}{13}=\frac{13+x}{13} }}\\\\\displaystyle{\sf{\longrightarrow 14\times13=13(13+x)}}\\\\\displaystyle{\sf{\longrightarrow 182=169+13x}}\\\\\displaystyle{\sf{\longrightarrow 196-182=13x}}\\\\\displaystyle{\sf{\longrightarrow 13=13x}}$

$\large\displaystyle{\sf{\color{lime}{\longrightarrow x=1 }}}$