Official solution
\[{T_{r + 1}} = {}^{10}{C_r}{(t{x^{1/5}})^{10 - r}}{\left[ {{{{{(1 - x)}^{1/10}}} \over t}} \right]^r}\]
\[ = {}^{10}{C_r}{t^{(10 - 2r)}} \times {x^{{{10 - r} \over 5}}} \times {(1 - x)^{{r \over {10}}}}\]
\[ \Rightarrow 10 - 2r = 0 \Rightarrow r = 5\]
\[ \therefore \] \[{T_6} = {}^{10}{C_5} \times x\sqrt {1 - x} \]
At maximum, \[{{d{T_6}} \over {dx}} = {}^{10}{C_5}\left[ {\sqrt {1 - x} - {x \over {2\sqrt {1 - x} }}} \right] = 0\]
\[ \Rightarrow \] \[ 1 - x = x/2 \Rightarrow 3x = 2 \Rightarrow x = 2/3\]
\[{T_6}{|_{\max }} = {{10!} \over {5!5!}} \times {2 \over {3\sqrt 3 }}\] = \[{{2.10!} \over {3\sqrt 3 {{(5!)}^2}}}\]