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(C) The general term $T_{r+1}$ in the expansion of $\left(\sqrt{x}-\frac{k}{x^{2}}\right)^{10}$ is given by:$T_{r+1} = {}^{10}C_{r} (x^{1/2})^{10-r} (

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$3$

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(C) The general term $T_{r+1}$ in the expansion of $\left(\sqrt{x}-\frac{k}{x^{2}}\right)^{10}$ is given by:$T_{r+1} = {}^{10}C_{r} (x^{1/2})^{10-r} (-k x^{-2})^{r}$$T_{r+1} = {}^{10}C_{r} x^{(10-r)/2} (-k)^{r} x^{-2r}$$T_{r+1} = {}^{10}C_{r} (-k)^{r} x^{(10-5r)/2}$For the constant term,the exponent of $x$ must be $0$:$\frac{10-5r}{2} = 0$ $\Rightarrow 5r = 10$ $\Rightarrow r = 2$Substituting $r = 2$ into the expression for $T_{r+1}$:$T_{3} = {}^{10}C_{2} (-k)^{2} = 405$$45 \cdot k^{2} = 405$$k^{2} = \frac{405}{45} = 9$$|k| = \sqrt{9} = 3$

If the constant term in the binomial expansion of $\left(\sqrt{x}-\frac{k}{x^{2}}\right)^{10}$ is $405$,then $|k|$ equals

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