ГДЗ по алгебре 8 класс Макарычев Задание 541

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Год:2021
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Задание 541

\[\boxed{\text{541\ (541).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[\textbf{а)}\ 2x^{2} - 5x - 3 = 0\]

\[D = 25 + 24 = 49\]

\[x_{1,2} = \frac{5 \pm \sqrt{49}}{2 \cdot 2} = \frac{5 \pm 7}{4}\]

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

\[\textbf{б)}\ 3x^{2} - 8x + 5 = 0\]

\[D_{1} = 4^{2} - 3 \cdot 5 = 16 - 15 = 1\]

\[x_{1,2} = \frac{4 \pm \sqrt{1}}{3} = \frac{4 \pm 1}{3}\]

\[x_{1} = 1;\ \ x_{2} = \frac{5}{3} = 1\frac{2}{3}.\]

\[\textbf{в)}\ 5x^{2} + 9x + 4 = 0\]

\[D = 81 - 80 = 1\]

\[x_{1,2} = \frac{- 9 \pm \sqrt{1}}{5 \cdot 2} = \frac{- 9 \pm 1}{10}\]

\[x_{1} = - \frac{10}{10} = - 1;\ \ x_{2} = - \frac{8}{10} = - 0,8.\]

\[\textbf{г)}\ 36y^{2} - 12y + 1 = 0\]

\[D_{1} = 6^{2} - 36 = 36 - 36 = 0\]

\[y = \frac{6}{36} = \frac{1}{6}.\]

\[\textbf{д)}\ 3t^{2} - 3t + 1 = 0\]

\[D = 9 - 12 = - 3\]

\[D < 0 - корней\ нет.\]

\[\textbf{е)}\ x^{2} + 9x - 22 = 0\]

\[D = 81 + 88 = 169\]

\[x_{1,2} = \frac{- 9 \pm \sqrt{169}}{2} = \frac{- 9 \pm 13}{2}\]

\[x_{1} = - 11;\ \ x_{2} = 2.\]

\[\textbf{ж)}\ y^{2} - 12y + 32 = 0\]

\[D_{1} = 6^{2} - 32 = 36 - 32 = 4\]

\[y_{1,2} = 6 \pm 2\]

\[y_{1} = 8;\ \ y_{2} = 4.\]

\[\textbf{з)}\ 100x^{2} - 160x + 63 = 0\]

\[D_{1} = 80^{2} - 63 \cdot 100 =\]

\[= 6400 - 6300 = 100\]

\[x_{1,2} = \frac{80 \pm \sqrt{100}}{100} = \frac{80 \pm 10}{100}\]

\[x_{1} = 0,9;\ \ x_{2} = 0,7\ \]

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