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

For $x=\frac{5}{7}$,if $t_k$ is the first negative term in the expansion of $(1+x)^{7/5}$,then $t_1+t_2+\ldots+t_k=$

If $x = \frac{1 \cdot 3}{3 \cdot 6} + \frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9} + \frac{1 \cdot 3 \cdot 5 \cdot 7}{3 \cdot 6 \cdot 9 \cdot 12} + \ldots$ to infinite terms,then $9x^2 + 24x = $

When $|x|>3$,the coefficient of $\frac{1}{x^n}$ in the expansion of $x^{3/2}(3+x)^{1/2}$ is

$1+\frac{1}{3}+\frac{1 \times 3}{3 \times 6}+\frac{1 \times 3 \times 5}{3 \times 6 \times 9}+\ldots \text{ to } \infty =$

For $|x| < \frac{4}{3}$, the approximate value of $\frac{1}{(4-3 x)^{\frac{1}{2}}}$ is