Question

1) x/(x-1)+3x/(x^2-3x+2)=6/(x-2) 2) (x+1)/(x+)+10x/(x^2+x-6)

Answer 1

$1). \\ \frac{x}{x-1} + \frac{3x}{ x^{2} -3x+2} = \frac{6}{x-2} \\ x^{2} -3x+2=0 \\ a=1;b=-3;c=2 \\ D= b^{2} -4ac=9-4*2=9-8=1 \\ x_{1} = \frac{-b+ \sqrt{D} }{2a} = \frac{3+1}{2} = \frac{4}{2} =2 \\ x_{2} = \frac{-b- \sqrt{D} }{2a} = \frac{3-1}{2} = \frac{2}{2} =1 \\ \frac{x}{x-1} + \frac{3x}{(x-2)(x-1)} = \frac{6}{x-2} \\ \frac{x(x-2)}{(x-2)(x-1)} + \frac{3x}{(x-2)(x-1)} - \frac{6(x-1)}{(x-2)(x-1)} =0 \\ \frac{x(x-2)+3x-6(x-1)}{(x-2)(x-1)} =0 \\ \frac{ x^{2} -2x+3x-6x+6}{(x-2)(x-1)} =0 \\$$\frac{ x^{2} -5x+6}{(x-2)(x-1)} =0 \\ x^{2} -5x+6=0 \\ a=1;b=-5;c=6 \\ D= b^{2} -4ac=25-4*6=25-24=1 \\ x_{1} = \frac{-b+ \sqrt{D} }{2a} = \frac{5+1}{2} = \frac{6}{2} =3 \\ x_{2} = \frac{-b- \sqrt{D} }{2a} = \frac{5-1}{2} = \frac{4}{2} =2 \\ \frac{(x-3)(x-2)}{(x-2)(x-1)} =0 \\ \frac{(x-3)}{(x-1)} =0$