Find the term independent of x …

{tex}{\left( {\sqrt {{x \over 3}} + {3 \over {2{x^2}}}} \right)^{10}}{/tex}

{tex}{T_{r + 1}}{ = ^{10}}{C_r}{\left( {\sqrt {{x \over 3}} } \right)^{10 - r}}{\left( {{3 \over {2{x^2}}}} \right)^r}{/tex}

{tex}{ = ^{10}}{C_r}{\left( x \right)^{5 - {r \over 2} - 2r}}{\left( 3 \right)^{ - 5 + {r \over 2} + r}}{\left( 2 \right)^{ - r}}{/tex}

{tex}{ = ^{10}}{C_r}{\left( x \right)^{{{10 - 5r} \over 2}}}{\left( 3 \right)^{{{3r - 10} \over 2}}}{\left( 2 \right)^{ - r}}{/tex}

{tex}for\,independent\,of\,x{/tex}

{tex}{\left( x \right)^{{{10 - 5r} \over 2}}} = {x^0}{/tex}

{tex}{{10 - 5r} \over 2} = 0{/tex}

{tex}10 - 5r = 0{/tex}

{tex}r = 2{/tex}

{tex}independent\,term\,{T_3}{ = ^{10}}{C_2}{\left( 3 \right)^{ - 2}}{\left( 2 \right)^{ - 2}}{/tex}

{tex} = {{10!} \over {2!8!}} \times {1 \over {36}}{/tex}

{tex} = {{10 \times 9} \over {2 \times 36}}{/tex}

{tex} = {5 \over 4}{/tex}