If $x > \sqrt{3}$ and $\frac{x^2+1}{(x^2+2)(x^2+3)}$ is expanded in terms of powers of $x^{-2}$,then the coefficient of $x^{-8}$ is

If the value of $x$ is so small that $x^2$ and higher powers can be neglected,then $\frac{\sqrt{1 + x} + \sqrt[3]{(1 - x)^2}}{1 + x + \sqrt{1 + x}}$ is equal to

Assertion $(A) : 1+\frac{2}{3} \cdot \frac{1}{2}+\frac{2 \cdot 5}{3 \cdot 6} \cdot \frac{1}{4}+\frac{2 \cdot 5 \cdot 8}{3 \cdot 6 \cdot 9} \cdot \frac{1}{8}+\ldots \infty = \sqrt[3]{4}$
Reason $(R) : |x| < 1, (1-x)^{-n} = 1+nx+\frac{n(n+1)}{1 \cdot 2} x^2+\frac{n(n+1)(n+2)}{1 \cdot 2 \cdot 3} x^3+\ldots$ The correct answer is

Assuming $|x|$ to be so small that $x^2$ and higher powers of $x$ can be neglected,then $\frac{\sqrt{1+x}+(1-x)^{3/2}}{(1+x)+\sqrt{1+x}} = $

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

When $|x| < \frac{1}{2}$,the coefficient of $x^6$ in the expansion of $\left(\frac{2-x}{1+2x}\right)^2$ is