The expression $\frac{1}{(x^2 + \frac{1}{x})^{4/3}}$ can be expanded by the binomial theorem if:

If $|x| < \frac{1}{2}$,then the coefficient of $x^r$ in the expansion of $\frac{1+2x}{(1-2x)^2}$ is

If $|x|$ is so small that $x^2$ and higher powers of $x$ may be neglected, then the approximate value of $\frac{\sqrt{4+x}+\sqrt[3]{8-x}}{\left(1-\frac{2x}{3}\right)^{\frac{3}{2}}}$ when $x=\frac{6}{25}$ is

If $|x| < 1$,then the value of $1 + n\left( \frac{2x}{1 + x} \right) + \frac{n(n + 1)}{2!}\left( \frac{2x}{1 + x} \right)^2 + \dots \infty$ will be

If the set of all values of $x$ for which the expansion of $(7-5 x)^{-\frac{2}{3}}$ is valid is equal to $(-a, a)$,then $5 a+7$ is equal to

The binomial expansion $(7+3x)^{-2/5}$ is valid for all $x$ in the interval $\left(\frac{-7}{3}, \frac{7}{3}\right)$. If the $4^{th}$ term of its expansion is $kx^3$,then the value of $(7^{12/5}k)$ is: