Страница 196 номер 688, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

1) \(3x^{2} - 4x + 1 = 3\left( x - \frac{1}{3} \right)(x - 1)\)
\(\dfrac{D}{4} = 4 - 3 = 1\)
\(x_{1} = \dfrac{2 - 1}{3} = \dfrac{1}{3};\ x_{2} = \dfrac{2 + 1}{3} = 1\)
\(\dfrac{9x^{2} - 1}{3x^{2} - 4x + 1} = \dfrac{(3x - 1)(3x + 1)}{3\left( x - \dfrac{1}{3} \right)(x - 1)} = \dfrac{3\left( x - \frac{1}{3} \right)(3x + 1)}{3\left( x - \dfrac{1}{3} \right)(x - 1)} = \dfrac{3x + 1}{x - 1}\)

2) \(2x^{2} - 5x + 3 = 2(x - 1)\left( x - \frac{3}{2} \right)\)
\(D = 25 - 4 \cdot 2 \cdot 3 = 25 - 24 = 1\)
\(x_{1} = \dfrac{5 - 1}{4} = 1;\ x_{2} = \dfrac{5 + 1}{4} = \dfrac{6}{4} = \dfrac{3}{2}\)
\(\dfrac{2x^{2} - 5x + 3}{4x^{2} - 12x + 9} = \dfrac{2(x - 1)\left( x - \frac{3}{2} \right)}{(2x - 3)^{2}} = \dfrac{(x - 1)(2x - 3)}{(2x - 3)^{2}} = \dfrac{x - 1}{2x - 3}\)

Ответ: 1) \(\frac{3x + 1}{x - 1}\); 2) \(\frac{x - 1}{2x - 3}\).