If the coefficient of {x^7} in {left( {a{x^2} | vedclass

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(A) In the expansion of ${\left( {a{x^2} + \frac{1}{{bx}}} \right)^{11}}$,the general term is ${T_{r + 1}} = {}^{11}{C_r}{(a{x^2})^{11 - r}}{\left( {\

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(A) In the expansion of ${\left( {a{x^2} + \frac{1}{{bx}}} \right)^{11}}$,the general term is ${T_{r + 1}} = {}^{11}{C_r}{(a{x^2})^{11 - r}}{\left( {\frac{1}{{bx}}} \right)^r} = {}^{11}{C_r}{a^{11 - r}}{b^{ - r}}{x^{22 - 3r}}$.For $x^7$,we set $22 - 3r = 7$,which gives $r = 5$. The coefficient is ${}^{11}{C_5}{a^6}{b^{ - 5}}$.In the expansion of ${\left( {ax - \frac{1}{{b{x^2}}}} \right)^{11}}$,the general term is ${T_{r + 1}} = {}^{11}{C_r}{(ax)^{11 - r}}{\left( { - \frac{1}{{b{x^2}}}} \right)^r} = {}^{11}{C_r}{( - 1)^r}{a^{11 - r}}{b^{ - r}}{x^{11 - 3r}}$.For $x^{ - 7}$,we set $11 - 3r = -7$,which gives $r = 6$. The coefficient is ${}^{11}{C_6}{( - 1)^6}{a^5}{b^{ - 6}} = {}^{11}{C_5}{a^5}{b^{ - 6}}$.Equating the coefficients: ${}^{11}{C_5}{a^6}{b^{ - 5}} = {}^{11}{C_5}{a^5}{b^{ - 6}}$.Dividing by ${}^{11}{C_5}{a^5}{b^{ - 5}}$,we get $a/1 = 1/b$,which implies $ab = 1$.

If the coefficient of ${x^7}$ in ${\left( {a{x^2} + \frac{1}{{bx}}} \right)^{11}}$ is equal to the coefficient of ${x^{ - 7}}$ in ${\left( {ax - \frac{1}{{b{x^2}}}} \right)^{11}}$,then $ab =$

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