Решебник по алгебре 9 класс Мерзляк Задание 604

\[\boxed{\mathbf{604\ (604).\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

\[1)\ y = \frac{3x^{2} - 10x + 3}{x - 3} - \frac{x^{2} - 1}{x + 1}\]

\[3x² - 10x + 3 = 0\]

\[D = 100 - 36 = 64\]

\[x_{1,2} = \frac{10 \pm 8}{6},\ \ x_{1} = 3,\ \ \]

\[x_{2} = \frac{1}{3}\]

\[\frac{3 \cdot (x - 3)\left( x - \frac{1}{3} \right)}{(x - 3)} - \frac{x^{2} - 1}{x + 1} =\]

\[= 3 \cdot \left( x - \frac{1}{3} \right) - \frac{(x - 1)(x + 1)}{(x + 1)} =\]

\[= 3 \cdot \left( x - \frac{1}{3} \right) - x + 1 =\]

\[= 3x - 1 - x + 1 = 2x\]

\[y = 2x,\ \ x \neq 3,\ \ x \neq - 1\]

\[2)\ y = \frac{5x^{2} + 4x - 1}{x + 1} - \frac{x^{2} - 3x}{x}\]

\[5x^{2} + 4x - 1 = 0\]

\[D = 16 + 20 = 36\]

\[x_{1,2} = \frac{- 4 \pm 6}{10},\ \ x_{1} = - 1,\]

\[\text{\ \ }x_{2} = \frac{1}{5}\]

\[\frac{5x^{2} + 4x - 1}{x + 1} - \frac{x^{2} - 3x}{x} =\]

\[= \frac{5 \cdot (x + 1)\left( x - \frac{1}{5} \right)}{(x + 1)} -\]

\[- \frac{x(x - 3)}{x} = 5x - 1 - x + 3 =\]

\[= 4x + 2\]

\[y = 4x + 2,\ \ x \neq 0,\ \]

\[\ x \neq - 1\]