[tex]\displaystyle\bf\\-\frac{6x+1}{16x^{2} -9} +\frac{x-2}{3-4x} =\frac{3x-1}{4x+3} \\\\\\\frac{6x+1}{(3-4x)(3+4x)} +\frac{x-2}{3-4x}-\frac{3x-1}{3+4x}=0\\\\\\\frac{6x+1+(x-2)\cdot(3+4x)-(3x-1)\cdot(3-4x)}{(3-4x)(3+4x)} =0\\\\\\\frac{6x+1+3x+4x^{2} -6-8x-9x+12x^{2} +3-4x}{(3-4x)(3+4x)} =0\\\\\\\frac{16x^{2} -12x-2}{(3-4x)(3+4x)} =0\\\\\\\left \{ {{16x^{2} -12x-2=0} \atop {3-4x\neq 0 \ ; \ 3+4x\neq 0}} \right.[/tex]

[tex]\displaystyle\bf\\16x^{2} -12x-2=0\\\\8x^{2} -6x-1=0\\\\D=(-6)^{2} -4\cdot 8\cdot(-1)=36+32=68=(2\sqrt{17})^{2} \\\\\\x_{1} =\frac{6-2\sqrt{17} }{16}=\frac{3-\sqrt{17} }{8} \\\\\\x_{2} =\frac{6+2\sqrt{17} }{16}=\frac{3+\sqrt{17} }{8}[/tex]

Найдём среднее арифметическое корней уравнения , для этого сложим их и разделим на 2 :

[tex]\displaystyle\bf\\\Big(x_{1} +x_{2} \Big):2=\Big(\frac{3-\sqrt{17} }{8} +\frac{3+\sqrt{17} }{8} \Big):2=\frac{3-\sqrt{17}+3+\sqrt{17} }{8} \cdot\frac{1}{2} =\\\\\\=\frac{6}{16}=0,375\\\\\\Otvet: \ 0,375[/tex]