ГДЗ по алгебре 9 класс Макарычев ФГОС Задание 163

\[\boxed{\text{164.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[\frac{(1 - 3a)^{2}}{3a^{2} + 5a - 2}\]

\[3a^{2} + 5a - 2 = 0\]

\[D = 5^{2} + 4 \cdot 3 \cdot 2 = 25 + 24 =\]

\[= 49\]

\[a_{1,2} = \frac{- 5 \pm 7}{6},\]

\[\text{\ \ }a_{1,2} = - 2;\frac{1}{3}.\]

\[\Longrightarrow 3a^{2} + 5a - 2 =\]

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

\[= (a + 2)(3a - 1);\]

\[\Longrightarrow \frac{(1 - 3a)^{2}}{(a + 2)(3a - 1)} =\]

\[= - \frac{1 - 3a}{a + 2} = \frac{3a - 1}{a + 2}.\]