помогите решить!!))

$\frac{2x^2-1}{x^2-9}-\frac{x+1}{x+3}=\frac{3x+1}{3x-9},\ |\bullet3\\\\\frac{6x^2-3}{(x-3)(x+3)}-\frac{3x+3}{x+3}-\frac{3x+1}{x-3}=0,\\\\\frac{6x^2-3-3(x+1)(x-3)-(3x+1)(x+3)}{(x-3)(x+3)}=0,$

Область допускаемых значений (ОДЗ): $(x-3)(x+3)\ne0,\\x\ne3,\ x\ne-3.$

$6x^2-3-3(x+1)(x-3)-(3x+1)(x+3)=0,\\6x^2-3-3(x^2-3x+x-3)-(3x^2+9x+x+3)=0,\\6x^2-3-3(x^2-2x-3)-(3x^2+10x+3)=0,\\6x^2-3-3x^2+6x+9-3x^2-10x-3=0,\\-4x+3=0,\\x= \frac{3}{4}.$