Question

Решите дробно-рациональное уравнение

Answer 1

$\bf \frac{ {x}^{2} - x }{ {x}^{2} - 6x + 9 } - \frac{1}{3} = \frac{3 - x}{3x - 9} ; \\ \bf {x }^{2} - 6x + 9; \\ \bf \: \: D = {b}^{2} - 4ac = 36 - 36 = 0\Rightarrow \: x1 = x2 = - \frac{b}{2a} = \frac{6}{2} = 3 ; \: \\ \bf a {x}^{2} + bx + c = a(x - x1)(x - x2) \Rightarrow \: {x}^{2} - 6x + 9 = (x - 3)(x - 3); \\ \: \underbrace{ \bf \frac{ {x}^{2} - x }{ {(x - 3)}^{2} } } _{ \bf \: 3} - \underbrace{ \bf \: \frac{1}{3} } _{ \bf \: {(x - 3)}^{2} } = \underbrace{ \bf \: \frac{3 - x}{ \bf \: 3(x - 3)} } _{ \bf \: x - 3}; \\ \bf \: 3 {x}^{2} - 3x - {(x - 3)}^{2} = (3 - x)(x - 3); \\ \bf \: 3 {x}^{2} - 3x - ( {x}^{2} - 6x + 9) = 3x - 9 - {x}^{2} + 3x; \\ \bf \: 3 {x}^{2} - 3x = 0; \\ \bf \: 3x(x - 1) = 0; \\ \bf \: x1 = 0; \: \: \: \: \: \: \: \: \: \: \: \: x2 = 1. \\ \bf \: ODZ: \: \: x \neq \: 3; \\ \bf \huge \: otvet: \: \: \: \: \: x1 = 0; \: \: \: \: \: \: x2 = 1.$