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МатематикаРешебник по алгебре 8 класс Макарычев ФГОС Задание 749
Задание 749
\[\boxed{\text{749\ (749).\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Знаки сравнения:
\(> \ - \ \)больше;
\(\mathbf{<} -\) меньше.
Положительные числа – это числа, которые больше нуля.
Отрицательные числа – это числа, которые меньше нуля.
Любое отрицательное число (со знаком « – ») всегда меньше любого положительного.
Решение.
\[\textbf{а)}\ a - 3 > b - 3\ \ \ и\ \ b > 4\]
\[\Longrightarrow a > b,\ \ b > 4\]
\[\Longrightarrow a > 0,\ \ b > 0\]
\[\textbf{б)}\ a - 8 > b - 8\ \ \ и\ \ \ a < - 12\]
\[\Longrightarrow a > b,\ \ a < - 12\]
\[\Longrightarrow a < 0,\ \ b < 0\]
\[\textbf{в)}\ 7a > 7b\ \ \ и\ \ \ b > \frac{1}{2}\]
\[\Longrightarrow a > b,\ \ b > \frac{1}{2}\]
\[\Longrightarrow a > 0,\ \ b > 0\]
\[\textbf{г)} - 2a > - 2b\ \ \ и\ \ \ b < - \frac{1}{3}\]
\[\Longrightarrow a < b,\ \ b < - \frac{1}{3}\]
\[\Longrightarrow a < 0,\ \ b < 0\]
\[\boxed{\text{749.}\text{\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\]
\[\textbf{а)}\ x^{2} - 2x - 5 = 0\]
\[D = 4 + 20 = 24\]
\[x_{1,2} = \frac{2 \pm \sqrt{24}}{2} = \frac{2 \pm 2\sqrt{6}}{2}\]
\[x_{1} = 1 + \sqrt{6};\ \ x_{2} = 1 - \sqrt{6}\]
\[Проверка:\]
\[\left( 1 + \sqrt{6} \right)^{2} - 2 \cdot \left( 1 + \sqrt{6} \right) - 5 =\]
\[= 1 + 2\sqrt{6} + 6 - 2 - 2\sqrt{6} - 5 =\]
\[= 0,\ \ 0 = 0.\]
\[\left( 1 - \sqrt{6} \right)^{2} - 2 \cdot \left( 1 - \sqrt{6} \right) - 5 =\]
\[= 1 - 2\sqrt{6} + 6 - 2 + 2\sqrt{6} - 5 =\]
\[= 0,\ \ 0 = 0.\]
\[\textbf{б)}\ x^{2} + 4x + 1 = 0\]
\[D = 16 + 4 = 20\]
\[x_{1,2} = \frac{- 4 \pm \sqrt{20}}{2} = \frac{- 4 \pm 2\sqrt{5}}{2}\]
\[x_{1} = - 2 + \sqrt{5};\ \ x_{2} = - 2 - \sqrt{5}\]
\[Проверка:\]
\[\left( - 2 + \sqrt{5} \right)^{2} + 4 \cdot \left( - 2 + \sqrt{5} \right) +\]
\[+ 1 = 4 - 4\sqrt{5} + 5 - 8 + 4\sqrt{5} +\]
\[+ 1 = 0,\ 0 = 0.\]
\[\left( - 2 - \sqrt{5} \right)^{2} + 4 \cdot \left( - 2 - \sqrt{5} \right) +\]
\[+ 1 = 4 + 4\sqrt{5} + 5 - 8 - 4\sqrt{5} +\]
\[+ 1 = 0,\ 0 = 0.\]
\[\textbf{в)}\ 3y^{2} - 4y - 2 = 0\]
\[D = 16 + 24 = 40\]
\[y_{1,2} = \frac{4 \pm \sqrt{40}}{6} = \frac{2 \pm \sqrt{10}}{3}\]
\[Проверка:\]
\[\left( \frac{2 + \sqrt{10}}{3}\ \right)^{2} \cdot 3 -\]
\[- \left( \frac{2 + \sqrt{10}}{3} \right) \cdot 4 - 2 =\]
\[= \frac{4 + 4\sqrt{10} + 10}{9} \cdot 3 -\]
\[- \frac{8 + 4\sqrt{10}}{3} - 2 = \ \]
\[= \frac{4 + 4\sqrt{10}}{3} - \frac{8 + 4\sqrt{10}}{3} - 2 =\]
\[= \frac{14 + 4\sqrt{10} - 8 - 4\sqrt{10} - 6}{3} =\]
\[= \frac{0}{3} = 0,\ \ 0 = 0\]
\[\left( \frac{2 - \sqrt{10}}{3} \right)^{2} \cdot 3 -\]
\[- \left( \frac{2 - \sqrt{10}}{3} \right) \cdot 4 -\]
\[- 2 = \frac{4 - 4\sqrt{10} + 10}{9} \cdot 3 -\]
\[- \frac{8 - 4\sqrt{10}}{3} - 2 =\]
\[= \frac{14 - 4\sqrt{10}}{3} - \frac{8 - 4\sqrt{10}}{3} - \frac{6}{3} =\]
\[= \frac{14 - 4\sqrt{10} - 8 + 4\sqrt{10} - 6}{3} =\]
\[= \frac{0}{3} = 0,\ 0 = 0.\]
\[\textbf{г)}\ 5y^{2} - 7y + 1 = 0\]
\[D = 49 - 20 = 29\]
\[y_{1,2} = \frac{7 \pm \sqrt{29}}{10}\]
\[Проверка:\]
\[5 \cdot \left( \frac{7 + \sqrt{29}}{10} \right)^{2} - 7 \cdot\]
\[\cdot \left( \frac{7 + \sqrt{29}}{10} \right) + 1 = 5 \cdot\]
\[\cdot \left( \frac{49 + 14\sqrt{29} + 29}{100} \right) -\]
\[- \frac{49 + 7\sqrt{29}}{10} + 1 =\]
\[= \frac{78 + 14\sqrt{29}}{20} -\]
\[- \frac{49 + 7\sqrt{29}}{10} + 1 =\]
\[= \frac{39 + 7\sqrt{29} - 49 - 7\sqrt{29} + 10}{10} =\]
\[= \frac{0}{10} = 0,\ \ 0 = 0.\]
\[5 \cdot \left( \frac{7 - \sqrt{29}}{10} \right)^{2} - 7 \cdot\]
\[\cdot \left( \frac{7 - \sqrt{29}}{10} \right) + 1 =\]
\[= 5 \cdot \left( \frac{49 - 14\sqrt{29} + 29}{100} \right) -\]
\[- \frac{49 - 7\sqrt{29}}{10} + 1 =\]
\[= \frac{78 - 14\sqrt{29}}{20} -\]
\[- \frac{49 - 7\sqrt{29}}{10} + 1 =\]
\[= \frac{39 - 7\sqrt{29} - 49 + 7\sqrt{29} + 10}{10} =\]
\[= \frac{0}{10} = 0,\ \ 0 = 0.\]
\[\textbf{д)}\ 2y^{2} + 11y + 10 = 0\]
\[D = 121 - 80 = 41\]
\[y_{1,2} = \frac{- 11 \pm \sqrt{41}}{4}\]
\[Проверка:\]
\[2 \cdot \left( \frac{- 11 + \sqrt{41}}{4} \right)^{2} + 11 \cdot\]
\[\cdot \left( \frac{- 11 + \sqrt{41}}{4} \right) + 10 = 2 \cdot\]
\[\cdot \left( \frac{121 - 22\sqrt{41} + 41}{16} \right) +\]
\[+ \frac{- 121 + 11\sqrt{41}}{4} + 10 =\]
\[= \frac{162 - 22\sqrt{41}}{8} +\]
\[+ \frac{- 121 + 11\sqrt{41}}{4} + \frac{40}{4} =\]
\[= \frac{81 - 11\sqrt{41} - 121 + 11\sqrt{41} + 40}{4} =\]
\[= \frac{0}{4} = 0,\ \ 0 = 0\]
\[2 \cdot \left( \frac{- 11 - \sqrt{41}}{4} \right)^{2} + 11 \cdot\]
\[\cdot \left( \frac{- 11 - \sqrt{41}}{4} \right) + 10 = 2 \cdot\]
\[\cdot \left( \frac{121 + 22\sqrt{41} + 41}{16} \right) +\]
\[+ \frac{- 121 - 11\sqrt{41}}{4} + 10 =\]
\[= \frac{162 + 22\sqrt{41}}{8} +\]
\[+ \frac{- 121 - 11\sqrt{41}}{4} + \frac{40}{4} =\]
\[= \frac{81 + 11\sqrt{41} - 121 - 11\sqrt{41} + 40}{4} =\]
\[= \frac{0}{4} = 0,\ \ 0 = 0\]
\[\textbf{е)}\ 4x^{2} - 9x - 2 = 0\]
\[D = 81 + 32 = 113\]
\[x_{1,2} = \frac{9 \pm \sqrt{113}}{8}\]
\[Проверка:\]
\[4 \cdot \left( \frac{9 + \sqrt{113}}{8} \right)^{2} - 9 \cdot\]
\[\cdot \left( \frac{9 + \sqrt{113}}{8} \right) - 2 = 4 \cdot\]
\[\cdot \left( \frac{81 + 18\sqrt{113} + 113}{64} \right) -\]
\[- \frac{81 + 9\sqrt{113}}{8} - 2 =\]
\[= \frac{194 + 18\sqrt{113}}{16} -\]
\[- \frac{81 + 9\sqrt{113}}{8} - \frac{16}{8} =\]
\[= \frac{97 + 9\sqrt{113} - 81 - 9\sqrt{113} - 16}{8} =\]
\[= \frac{0}{8} = 0,\ \ 0 = 0.\]
\[4 \cdot \left( \frac{9 - \sqrt{113}}{8} \right)^{2} - 9 \cdot\]
\[\cdot \left( \frac{9 - \sqrt{113}}{8} \right) - 2 = 4 \cdot\]
\[\cdot \left( \frac{81 - 18\sqrt{113} + 113}{64} \right) -\]
\[- \frac{81 - 9\sqrt{113}}{8} - 2 =\]
\[= \frac{194 - 18\sqrt{113}}{16} -\]
\[- \frac{81 - 9\sqrt{113}}{8} - \frac{16}{8} =\]
\[= \frac{97 - 9\sqrt{113} - 81 + 9\sqrt{113} - 16}{8} =\]
\[= \frac{0}{8} = 0,\ \ 0 = 0.\]
\[\ \]
