ГДЗ по алгебре 8 класс Макарычев Задание 539

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Год:2021
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Задание 539

Содержание

\[\boxed{\text{539\ (539).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 3x^{2} - 14x + 16 = 0\]

\[D_{1} = 7^{2} - 3 \cdot 16 = 49 - 48 = 1\]

\[x_{1,2} = \frac{7 \pm \sqrt{1}}{3} = \frac{7 \pm 1}{3}\]

\[x_{1} = \frac{8}{3} = 2\frac{2}{3};\ \ x_{2} = \frac{6}{3} = 2\]

\[Ответ:x = 2\frac{2}{3};\ \ x = 2.\]

\[\textbf{б)}\ 5x^{2} - 16x + 3 = 0\]

\[D_{1} = 8^{2} - 5 \cdot 3 = 64 - 15 = 49\]

\[x_{1,2} = \frac{8 \pm \sqrt{49}}{5} = \frac{8 \pm 7}{5}\]

\[x_{1} = \frac{15}{5} = 3;\ \ x_{2} = \frac{1}{5} = 0,2\]

\[Ответ:x = 3;\ \ x = 0,2.\]

\[\textbf{в)}\ x^{2} + 2x - 80 = 0\]

\[D_{1} = 1 + 80 = 81\]

\[x_{1,2} = \frac{- 1 \pm \sqrt{81}}{1} = - 1 \pm 9\]

\[x_{1} = - 10;\ \ x_{2} = 8\]

\[Ответ:x = - 10;\ \ x = 8.\]

\[\textbf{г)}\ x^{2} - 22x - 23 = 0\]

\[D_{1} = 11^{2} + 23 \cdot 1 = 121 + 23 =\]

\[= 144\]

\[x_{1,2} = \frac{11 \pm \sqrt{144}}{1} = 11 \pm 12\]

\[x_{1} = 23;\ \ x_{2} = - 1\]

\[Ответ:x = 23;\ \ x = - 1.\]

\[\textbf{д)}\ 4x^{2} - 36x + 77 = 0\]

\[D_{1} = 18^{2} - 4 \cdot 77 =\]

\[= 324 - 308 = 16\]

\[x_{1,2} = \frac{18 \pm \sqrt{16}}{4} = \frac{18 \pm 4}{4}\]

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

\[Ответ:x = 3,5;\ \ x = 5,5.\]

\[\textbf{е)}\ 15y^{2} - 22y - 37 = 0\]

\[D_{1} = 11^{2} + 15 \cdot 37 =\]

\[= 121 + 555 = 676\]

\[y_{1,2} = \frac{11 \pm \sqrt{676}}{15} = \frac{11 \pm 26}{15}\]

\[y_{1} = - 1;\ \ y_{2} = \frac{37}{15} = 2\frac{7}{15}\]

\[Ответ:y = - 1;\ \ y = 2\frac{7}{15}.\]

\[\textbf{ж)}\ 7z^{2} - 20z + 14 = 0\]

\[D_{1} = 10^{2} - 7 \cdot 14 = 100 - 98 =\]

\[= 2\]

\[z_{1,2} = \frac{10 \pm \sqrt{2}}{7}\]

\[z_{1} = \frac{10 + \sqrt{2}}{7};\ \ z_{2} = \frac{10 - \sqrt{2}}{7}\]

\[Ответ:z = \frac{10 \pm \sqrt{2}}{7}.\]

\[\textbf{з)}\ y^{2} - 10y - 25 = 0\]

\[D_{1} = 5^{2} + 25 \cdot 1 = 25 + 25 = 50\]

\[y_{1,2} = \frac{5 \pm \sqrt{50}}{1}\]

\[y_{1} = 5 + 5\sqrt{2};\ \ y_{2} = 5 - 5\sqrt{2}\text{\ \ \ }\]

\[Ответ:y = 5 \pm 5\sqrt{2}.\]

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