Решебник по алгебре 9 класс Макарычев Задание 133

 

\[\boxed{\text{133\ (133).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ (x - 1)^{2} + (x + 1)^{2} =\]

\[= (x + 2)^{2} - 2x + 2\]

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

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

\[2x^{2} + 2 = x^{2} + 2x + 6\]

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

\[D_{1} = 1 + 4 = 5\]

\[x_{1,2} = 1 \pm \sqrt{5}.\]

\[Ответ:x = 1 \pm \sqrt{5}.\]

\[\textbf{б)}\ (2x - 3)(2x + 3) - 1 =\]

\(= 5x + (x - 2)^{2}\)

\[4x^{2} - 9 - 1 = 5x + x^{2} - 4x + 4\]

\[3x^{2} - x - 14 = 0\]

\[D = 1 + 4 \cdot 3 \cdot 14 = 169\]

\[x_{1} = \frac{1 - 13}{6} = - 2;\ \ \ \ x_{2} =\]

\[= \frac{1 + 13}{6} = \frac{14}{6} = 2\frac{1}{3}.\]

\[Ответ:x = - 2;\ \ x = 2\frac{1}{3}.\]

\[\boxed{\text{133.}\text{\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[(x + 3)^{2} - (x - 3)^{2} =\]

\[= (x - 2)^{2} + (x + 2)^{2}\]

\[x^{2} + 6x + 9 - x^{2} + 6x - 9 =\]

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

\[12x = 2x^{2} + 8\]

\[x^{2} - 6x + 4 = 0\]

\[D = 3^{2} - 4 = 9 - 4 = 5\]

\[x_{1,2} = 3 \pm \sqrt{5}\text{\ \ \ }\]