Question

Боже с рациональным уравнением умераю

Answer 1

$\displaystyle\bf\\\frac{x^{2} -x}{x^{2} -6x+9} -\frac{1}{3} =\frac{3-x}{3x-9} \\\\\\\frac{x^{2} -x}{(x-3)^{2} } -\frac{1}{3} -\frac{3-x}{3(x-3)}=0 \\\\\\\frac{(x^{2} -x)\cdot 3-(x^{2} -6x+9)-(3-x)\cdot(x-3)}{3(x-3)^{2} } =0\\\\\\\frac{3x^{2} -3x-x^{2} +6x-9-3x+9+x^{2} -3x}{3((x-3)^{2} } =0\\\\\\\frac{3x^{2} -3x}{3(x-3)^{2} }=0\\\\\\\frac{3x(x-1)}{3(x-3)^{2} } =0\\\\\\\frac{x(x-1)}{(x-3)^{2} } =0\\\\\\\left \{ {{x(x-1)=0} \atop {x-3\neq }0} \right.$

$\displaystyle\bf\\\left \{ {{\left[\begin{array}{ccc}x_{1} =0\\x_{2} =1\end{array}\right } \atop {x\neq 3}} \right. \\\\\\Otvet:0 \ ; \ 1$

Answer 2

Подробное решение на фото