помогите решить

0;\\\\ x \neq 3; x \neq -1;\\\\ \frac{3}{x-2+\frac{3}{x}}+\frac{6}{x-3+\frac{3}{x}}=14;\\\\ " alt="\frac{3x}{x^2-2x+3}+\frac{6x}{x^2-3x+3}=14;\\\\ x^2-2x+3 \neq 0; x^2-3x+3 \neq 0;\\\\ (x-3)(x+1) \neq 0;x^2-3x+3=(x-1.5)^2+0.75>0;\\\\ x \neq 3; x \neq -1;\\\\ \frac{3}{x-2+\frac{3}{x}}+\frac{6}{x-3+\frac{3}{x}}=14;\\\\ " align="absmiddle" class="latex-formula">

$y=x+\frac{3}{x}-2.5;\\\\ \frac{3}{y+0.5}+\frac{6}{y-0.5}=14;\\\\ 3(y-0.5)+6(y+0.5)=14(y+0.5)(y-0.5);\\\\ 3y-1.5+6y+3=14y^2-3.5;\\\\\ 14y^2-9y-5=0;\\\\D=(-9)^2-4*14*(-5)=361=19^2;\\\\ y_1=\frac{9-19}{2*14}=-\frac{10}{28}=-\frac{5}{14};\\\\ y_2=\frac{9+19}{2*14}=\frac{28}{28}=1;\\\\$

$x+\frac{3}{x}-2.5=-\frac{5}{14};\\\\14x^2-35x+42+5x=0;\\\\14x^2-30x+42=0;\\\\7x^2-15x+21=0;\\\\D=(-15)^2-4*7*21=-363<0;\\\\x+\frac{3}{x}-2.5=1;\\\\x^2-3.5x+3=0;\\\\D=(-3.5)^2-4*1*3=0.25;\\\\x_1=\frac{3.5-0.25}{2*1}=1.625;\\\\x_2=\frac{3.5+0.25}{2*1}=1.875$