ĐK:\(\left\{ \begin{array}{l} x > 0\\ 6 + x - {x^2} > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x > 0\\ - 2 \le x \le 3 \end{array} \right.\)

Ta có:

\(\begin{array}{l} 5x + \sqrt {6{x^2} + {x^3} - {x^4}} {\log _2}x > \left( {{x^2} - x} \right){\log _2}x + 5 + 5\sqrt {6 + x - {x^2}} \\ \Leftrightarrow 5x + x\sqrt {6 + x - {x^2}} {\log _2}x > x\left( {x - 1} \right){\log _2}x + 5 + 5\sqrt {6 + x - {x^2}} \\ \Leftrightarrow \left( {x - 1} \right)\left( {5 - x{{\log }_2}x} \right) + \sqrt {6 + x - {x^2}} \left( {x{{\log }_2}x - 5} \right) > 0\\ \Leftrightarrow \left( {5 - x{{\log }_2}x} \right)\left( {x - 1 - \sqrt {6 + x - {x^2}} } \right) > 0 \end{array}\)

\( \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} 5 - x{\log _2}x > 0\\ x - 1 - \sqrt {6 + x - {x^2}} > 0 \end{array} \right.\\ \left\{ \begin{array}{l} 5 - x{\log _2}x < 0\\ x - 1 - \sqrt {6 + x - {x^2}} < 0 \end{array} \right. \end{array} \right.\)

Xét hệ \(\left( I \right)\left\{ \begin{array}{l} 5 - x{\log _2}x > 0{\rm{ }}\left( 1 \right)\\ x - 1 - \sqrt {6 + x - {x^2}} > 0{\rm{ }}\left( 2 \right) \end{array} \right.\)

Giải \(\left( 1 \right)\)

Xét hàm số \(f\left( x \right)=x\left( \frac{5}{x}-{{\log }_{2}}x \right)=xg\left( x \right)\) với \(x\in \left( 0;3 \right].\)

Ta có \(g'\left( x \right)=-\frac{5}{{{x}^{2}}}-\frac{1}{x\ln 2}<0,\forall x\in \left( 0;3 \right].\)

Lập bảng biến thiên:

Vậy \(f\left( x \right)=x\left( \frac{5}{x}-{{\log }_{2}}x \right)>0,\forall x\in \left( 0;3 \right].\)

Xét bất phương trình \(\left( 2 \right):\)

\(\left( 2 \right)\Leftrightarrow \sqrt{6+x-{{x}^{2}}}<x-1\)

\(\begin{array}{l} \Leftrightarrow \left\{ \begin{array}{l} 6 + x - {x^2} < {\left( {x - 1} \right)^2}\\ x > 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} {x^2} - 3x - 5 > 0\\ x > 1 \end{array} \right.\\ \Leftrightarrow \left\{ \begin{array}{l} \left[ \begin{array}{l} x < - 1\\ x > \frac{5}{2} \end{array} \right.\\ x > 1 \end{array} \right.\\ \Leftrightarrow x > \frac{5}{2}. \end{array}\)

Vậy nghiệm của hệ \(\left( I \right)\) là \(D=\left( \frac{5}{2};3 \right].\)

Hệ \(\left\{ \begin{array}{l} 5 - x{\log _2}x < 0\\ x - 1 - \sqrt {6 + x - {x^2}} < 0 \end{array} \right.\) vô nghiệm

Vậy \(S=\left( \frac{5}{2};3 \right],\) suy ra \(b-a=3-\frac{5}{2}=\frac{1}{2}.\)