ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 628

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Пояснение.

Чтобы убрать дроби, умножим обе части уравнение на НОК знаменателей.

Решение.

\[\textbf{а)}\ \frac{x^{2} - 1}{2} - 11x = 11\ \ \ \ \ | \cdot 2\]

\[x² - 1 - 22x = 22\]

\[x^{2} - 22x - 23 = 0\]

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

\[= 144\]

\[x_{1} = 11 + 12 = 23;\]

\[x_{2} = 11 - 12 = - 1.\]

\[Ответ:x = - 1;x = 23.\]

\[\textbf{б)}\ \frac{x^{2} + x}{2} = \frac{8x - 7}{3}\ \ \ \ \ \ | \cdot 6\]

\[3 \cdot \left( x^{2} + x \right) = 2 \cdot (8x - 7)\ \]

\[3x² + 3x = 16x - 14\]

\[3x^{2} - 13x + 14 = 0\]

\[D = 13^{2} - 4 \cdot 3 \cdot 14 =\]

\[= 169 - 168 = 1\]

\[x_{1} = \frac{13 + 1}{6} = \frac{14}{6} = \frac{7}{3} = 2\frac{1}{3};\]

\[x_{2} = \frac{13 - 1}{6} = 2.\]

\[Ответ:x = 2;\ \ x = 2\frac{1}{3}.\]

\[\textbf{в)}\ x - 3 = \frac{1 - x^{2}}{3}\ \ \ \ \ \ \ \ | \cdot 3\]

\[3 \cdot (x - 3) = 1 - x^{2}\]

\[3x - 9 - 1 + x^{2} = 0\]

\[x^{2} + 3x - 10 = 0\]

\[D = 9 + 40 = 49\]

\[x_{1} = \frac{- 3 + 7}{2} = 2;\ \ \ \ \ \]

\[x_{2} = \frac{- 3 - 7}{2} = - 5.\]

\[Ответ:x = - 5;\ \ x = 2.\]

\[\textbf{г)}\ \frac{2 - x^{2}}{7} = \frac{x}{2}\ \ \ \ \ \ \ \ | \cdot 14\]

\[2 \cdot \left( 2 - x^{2} \right) = 7x\]

\[4 - 2x^{2} - 7x = 0\ \ \ \ |\ :( - 1)\]

\[2x^{2} + 7x - 4 = 0\]

\[D = 49 + 32 = 81\]

\[x_{1} = \frac{- 7 + 9}{4} = \frac{2}{4} = 0,5;\ \ \ \ \ \]

\[x_{2} = \frac{- 7 - 9}{4} = - 4.\]

\[Ответ:x = - 4;\ \ x = 0,5.\]