The term independent of x in {left( {sqrt x - | vedclass

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(A) The general term in the expansion of ${\left( {\sqrt x - \frac{2}{x}} \right)^{18}}$ is given by $T_{r+1} = ^{18}C_r (\sqrt{x})^{18-r} \left( -\fr

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$^{18}C_6 	imes 2^6$

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(A) The general term in the expansion of ${\left( {\sqrt x - \frac{2}{x}} ight)^{18}}$ is given by $T_{r+1} = ^{18}C_r (\sqrt{x})^{18-r} \left( -\frac{2}{x} ight)^r$.Simplifying the expression,we get $T_{r+1} = ^{18}C_r (x^{1/2})^{18-r} (-2)^r (x^{-1})^r = ^{18}C_r x^{9 - r/2} (-2)^r x^{-r} = ^{18}C_r (-2)^r x^{9 - 3r/2}$.For the term to be independent of $x$,the exponent of $x$ must be $0$,so $9 - \frac{3r}{2} = 0$.Solving for $r$,we get $3r = 18$,which implies $r = 6$.Substituting $r = 6$ into the general term,the independent term is $T_7 = ^{18}C_6 (-2)^6 = ^{18}C_6 	imes 2^6$.

The term independent of $x$ in ${\left( {\sqrt x - \frac{2}{x}} \right)^{18}}$ is

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