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МатематикаГДЗ по алгебре 8 класс Макарычев ФГОС Задание 537
Задание 537
\[\boxed{\text{537\ (537).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
Пояснение.
Решение.
\[\textbf{а)}\ x^{2} - 11x + 31 = 1\]
\[x^{2} - 11x + 30 = 0\]
\[D = 121 - 120 = 1\]
\[x_{1,2} = \frac{11 \pm \sqrt{1}}{2} = \frac{11 \pm 1}{2}\]
\[x_{1} = 6;\ \ x_{2} = 5\]
\[Ответ:при\ x = 5;\ \ x = 6.\]
\[\textbf{б)}\ x^{2} - 5x - 3 = 2x - 5\]
\[x^{2} - 7x + 2 = 0\]
\[D = 49 - 8 = 41\]
\[x_{1,2} = \frac{7 \pm \sqrt{41}}{2}\]
\[x_{1} = \frac{7 - \sqrt{41}}{2};\ \ x_{2} = \frac{7 + \sqrt{41}}{2}\]
\[Ответ:при\ x = \frac{7 \pm \sqrt{41}}{2}.\]
\[\textbf{в)}\ 7x + 1 = 3x^{2} - 2x + 1\]
\[3x^{2} - 9x = 0\]
\[3x(x - 3) = 0\]
\[x_{1} = 0,\ \ \ \ x_{2} = 3\ \]
\[Ответ:при\ x = 0;\ \ x = 3.\]
\[\textbf{г)} - 2x^{2} + 5x + 6 = 4x^{2} + 5x\]
\[6x^{2} - 6 = 0\]
\[6 \cdot \left( x^{2} - 1 \right) = 0\]
\[x^{2} - 1 = 0\]
\[x^{2} = 1\]
\[x_{1,2} = \pm 1\ \ \]
\[Ответ:при\ x = \pm 1.\]
\(\boxed{\text{537.\ }\text{еуроки}\text{-}\text{ответы}\text{\ }\text{на}\text{\ }\text{пятёрку}}\)
Пояснение.
Решение.
\[\textbf{а)}\ 3x^{2} - 14x + 16 = 0\]
\[D_{1} = 49 - 48 = 1\]
\[x_{1,2} = \frac{7 \pm \sqrt{1}}{3} = \frac{7 \pm 1}{3}\]
\[x_{1} = \frac{8}{3} = 2\frac{2}{3};\ \ x_{2} = \frac{6}{3} = 2\]
\[Ответ:x = 2\frac{2}{3};\ \ x = 2.\]
\[\textbf{б)}\ 5x^{2} - 16x + 3 = 0\]
\[D_{1} = 64 - 15 = 49\]
\[x_{1,2} = \frac{8 \pm \sqrt{49}}{5} = \frac{8 \pm 7}{5}\]
\[x_{1} = \frac{15}{5} = 3;\ \ x_{2} = \frac{1}{5} = 0,2\]
\[Ответ:x = 3;\ \ x = 0,2.\]
\[\textbf{в)}\ x^{2} + 2x - 80 = 0\]
\[D_{1} = 1 + 80 = 81\]
\[x_{1,2} = \frac{- 1 \pm \sqrt{81}}{1} = - 1 \pm 9\]
\[x_{1} = - 10;\ \ x_{2} = 8\]
\[Ответ:x = - 10;\ \ x = 8.\]
\[\textbf{г)}\ x^{2} - 22x - 23 = 0\]
\[D_{1} = 121 + 23 = 144\]
\[x_{1,2} = \frac{11 \pm \sqrt{144}}{1} = 11 \pm 12\]
\[x_{1} = 23;\ \ x_{2} = - 1\]
\[Ответ:x = 23;\ \ x = - 1.\]
\[\textbf{д)}\ 4x^{2} - 36x + 77 = 0\]
\[D_{1} = 324 - 308 = 16\]
\[x_{1,2} = \frac{18 \pm \sqrt{16}}{4} = \frac{18 \pm 4}{4}\]
\[x_{1} = 5,5;\ \ x_{2} = 3,5\]
\[Ответ:x = 3,5;\ \ x = 5,5.\]
\[\textbf{е)}\ 15y^{2} - 22y - 37 = 0\]
\[D_{1} = 121 + 555 = 676\]
\[y_{1,2} = \frac{11 \pm \sqrt{676}}{15} = \frac{11 \pm 26}{15}\]
\[y_{1} = - 1;\ \ y_{2} = \frac{37}{15} = 2\frac{7}{15}\]
\[Ответ:y = - 1;\ \ y = 2\frac{7}{15}.\]
\[\textbf{ж)}\ 7z^{2} - 20z + 14 = 0\]
\[D_{1} = 100 - 98 = 2\]
\[z_{1,2} = \frac{10 \pm \sqrt{2}}{7}\]
\[z_{1} = \frac{10 + \sqrt{2}}{7};\ \ z_{2} = \frac{10 - \sqrt{2}}{7}\]
\[Ответ:z = \frac{10 \pm \sqrt{2}}{7}.\]
\[\textbf{з)}\ y^{2} - 10y - 25 = 0\]
\[D_{1} = 25 + 25 = 50\]
\[y_{1,2} = \frac{5 \pm \sqrt{50}}{1}\]
\[y_{1} = 5 + 5\sqrt{2};\ \ y_{2} = 5 - 5\sqrt{2}\text{\ \ \ }\]
\[Ответ:y = 5 \pm 5\sqrt{2}.\]
