Всем привет! Помогите пожалуйста

$\frac{6}{x^2-4x+3}-\frac{13-7x}{1-x}=\frac{3}{x-3}\\\\ \frac{6}{x^2-3x-x+3}-\frac{-(7x-13)}{-(x-1)}-\frac{3}{x-3}=0\\\\ \frac{6}{x(x-3)-(x-3)}-\frac{7x-13}{x-1}-\frac{3}{x-3}=0\\\\ \frac{6}{(x-1)(x-3)}-\frac{7x-13}{x-1}-\frac{3}{x-3}=0\\\\ \frac{6-(x-3)(7x-13)-3(x-1)}{(x-1)(x-3)}=0\\\\ \left \{ {{6-[7x^2-13x-21x+39]-3x+3=0} \atop {(x-1)(x-3)\neq0}} \right. \\\\ \left \{ {{9-3x-[7x^2-34x+39]=0} \atop {x-1\neq0\ \ or\ \ x-3\neq0}} \right. \\\\$

$\left \{ {{7x^2-34x+39-9+3x=0} \atop {x\neq1\ \ or\ \ x\neq3}} \right. \\\\ \left \{ {{7x^2-31x+30=0} \atop {x\neq1\ \ or\ \ x\neq3}} \right. \\\\ \left \{ {{7x^2-21x-10x+30=0} \atop {x\neq1\ \ or\ \ x\neq3}} \right. \\\\ \left \{ {{7x(x-3)-10(x-3)=0} \atop {x\neq1\ \ or\ \ x\neq3}} \right. \\\\ \left \{ {{(7x-10)(x-3)=0} \atop {x\neq1\ \ or\ \ x\neq3}} \right. \\\\ \left \{ {{7x-10=0\ or\ \ x-3=0} \atop {x\neq1\ \ or\ \ x\neq3}} \right. \\\\$

$\left \{ {{x=\frac{10}{7}\ or\ \ x=3} \atop {x\neq1\ \ or\ \ x\neq3}} \right. \\\\ x=\frac{10}{7}\\\\$

Ответ: $\frac{10}{7}$