Question

Решите дробно-рациональное уравнение

Answer 1

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

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