Решебник по алгебре 9 класс Макарычев Задание 60

 

\[\boxed{\text{60\ (60).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 10x^{2} + 5x - 5 = 0\]

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

\[D = 1 + 4 \cdot 2 \cdot 1 = 9\]

\[\ x_{1} = \frac{- 1 + 3}{4} = \frac{1}{2} = 0,5;\ \ \ x_{2} =\]

\[= \frac{- 1 - 3}{4} = - 1.\]

\[Ответ:x = 0,5;\ \ x = 1.\]

\[\textbf{б)} - 2x^{2} + 12x - 18 = 0\ |\ :( - 2)\]

\[x^{2} - 6x + 9 = 0\]

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

\[x - 3 = 0\]

\[x = 3\]

\[Ответ:x = 3.\]

\[\textbf{в)}\ x^{2} - 2x - 4 = 0\]

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

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

\[Ответ:x = 1 \pm \sqrt{5}.\]

\[\textbf{г)}\ 12x^{2} - 12 = 0\]

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

\[x^{2} = 1\]

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

\[Ответ:x = \pm 1.\]

\[\boxed{\text{60.\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]

\[x^{2} - x - 240 = 1\]

\[x_{1} + x_{2} = 1;\ \ \ x_{1} \cdot x_{2} = - 240\]

\[x_{1} = 16;\ \ \ x_{2} = - 15\]

\[y_{1} = 16 - 1 = 15;\]

\[y_{2} = - 15 - 1 = - 16.\]

\[Ответ:(16;15);\ \ ( - 15; - 16).\]

\[x^{2} + 4x^{2} - 60x + 225 - 65 = 0\]

\[5x^{2} - 60x + 160 = 0\ \ \ \ \ |\ :5\]

\[x^{2} - 12x + 32 = 0\]

\[D_{1} = 36 - 32 = 4\]

\[x_{1} = 6 + 2 = 8;\ \ \ x_{2} = 6 - 2 = 4.\]

\[y_{1} = 2 \cdot 8 - 15 = 1;\]

\[y_{2} = 2 \cdot 4 - 15 = - 7.\]

\[Ответ:(8;1);\ \ (4; - 7).\]