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If $(-c, c)$ is the set of all values of $x$ for which the expansion of $(7-5 x)^{-2 / 3}$ is valid, then $5 c+7=$
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1176 Upvotes
Verified Answer
The correct answer is:
$14$
$(7-5 x)^{\frac{-2}{3}}$
$$
=7^{\frac{-2}{3}}\left(1-\frac{5 x}{7}\right)^{-\frac{2}{3}}
$$
To make expansion valid
$$
\begin{aligned}
& \left|\frac{5 x}{7}\right| < 1 \Rightarrow|x| < \frac{7}{5} \\
& \therefore x \in\left(\frac{-7}{5}, \frac{7}{5}\right) \\
& \therefore C=\frac{7}{5} \therefore 5 C+7=14
\end{aligned}
$$
$$
=7^{\frac{-2}{3}}\left(1-\frac{5 x}{7}\right)^{-\frac{2}{3}}
$$
To make expansion valid
$$
\begin{aligned}
& \left|\frac{5 x}{7}\right| < 1 \Rightarrow|x| < \frac{7}{5} \\
& \therefore x \in\left(\frac{-7}{5}, \frac{7}{5}\right) \\
& \therefore C=\frac{7}{5} \therefore 5 C+7=14
\end{aligned}
$$
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