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МатематикаРешебник по алгебре 9 класс Макарычев Задание 87
Задание 87
\[\boxed{\text{87\ (}\text{н}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \frac{x^{2} - 1}{2} - 11x = 11\ \ \ \ \ | \cdot 2\]
\[x^{2} - 1 - 22x = 22\]
\[x^{2} - 22x - 23 = 0\]
\[D = 11^{2} + 23 = 121 + 23 = 144\]
\[x_{1} = 11 + 12 = 23;\ \ x_{2} =\]
\[= 11 - 12 = - 1.\]
\[Ответ:x = 23;x = - 1.\]
\[\textbf{б)}\frac{x^{2} + x}{2} = \frac{8x - 7}{3}\ \ \ \ \ \ | \cdot 6\]
\[3 \cdot \left( x^{2} + x \right) = 2 \cdot (8x - 7)\ \]
\[3x^{2} + 3x = 16x - 14\]
\[3x^{2} - 13x + 14 = 0\]
\[D = 13^{2} - 4 \cdot 3 \cdot 14 =\]
\[= 169 - 168 = 1\]
\[x_{1} = \frac{13 + 1}{6} = \frac{14}{6} = \frac{7}{3} =\]
\[= 2\frac{1}{3};\ \ x_{2} = \frac{13 - 1}{6} = 2.\]
\[Ответ:x = 2;\ \ x = 2\frac{1}{3}.\]
\[\textbf{в)}\ x - 3 = \frac{1 - x^{2}}{3}\ \ \ \ \ \ \ \ | \cdot 3\]
\[3 \cdot (x - 3) = 1 - x^{2}\]
\[3x - 9 - 1 + x^{2} = 0\]
\[x^{2} + 3x - 10 = 0\]
\[D = 9 + 40 = 49\]
\[x_{1} = \frac{- 3 + 7}{2} = 2;\ \ \ \ \ x_{2} =\]
\[= \frac{- 3 - 7}{2} = - 5\]
\[Ответ:x = 2;\ \ x = - 5.\]
\[\textbf{г)}\ \frac{2 - x^{2}}{7} = \frac{x}{2}\ \ \ \ \ \ \ \ | \cdot 14\]
\[2 \cdot \left( 2 - x^{2} \right) = 7x\]
\[4 - 2x^{2} - 7x = 0\ \ \ \ |\ :( - 1)\]
\[2x^{2} + 7x - 4 = 0\]
\[D = 49 + 32 = 81\]
\[x_{1} = \frac{- 7 + 9}{4} = \frac{2}{4} = 0,5;\ \ \ \ \ \]
\[x_{2} = \frac{- 7 - 9}{4} = - 4\]
\[Ответ:x = - 4;\ \ x = 0,5.\]
\[\boxed{\text{87\ (}\text{с}\text{).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[\textbf{а)}\ \frac{x^{2} - 1}{2} - 11x = 11\ \ \ \ \ | \cdot 2\]
\[x^{2} - 1 - 22x = 22\]
\[x^{2} - 22x - 23 = 0\]
\[D_{1} = 11^{2} + 23 = 121 + 23 = 144\]
\[x_{1} = 11 + 12 = 23;\ \ x_{2} =\]
\[= 11 - 12 = - 1.\]
\[Ответ:x = 23;x = - 1.\]
\[\textbf{б)}\ \frac{x^{2} + x}{2} = \frac{8x - 7}{3}\ \ \ \ \ \ | \cdot 6\]
\[3 \cdot \left( x^{2} + x \right) = 2 \cdot (8x - 7)\ \]
\[3x^{2} + 3x = 16x - 14\]
\[3x^{2} - 13x + 14 = 0\]
\[D = 13^{2} - 4 \cdot 3 \cdot 14 =\]
\[= 169 - 168 = 1\]
\[x_{1} = \frac{13 + 1}{6} = \frac{14}{6} = \frac{7}{3} = 2\frac{1}{3};\ \ \]
\[x_{2} = \frac{13 - 1}{6} = 2.\]
\[Ответ:x = 2;\ \ x = 2\frac{1}{3}.\]
\[\boxed{\text{87.\ }\text{ОК\ ГДЗ\ -\ домашка\ на\ 5}}\]
\[\textbf{а)}\ \frac{2a + 5b}{5a + 2b} = 1\]
\[2a + 5b = 5a + 2b\]
\[2a - 5a = 2b - 5b\]
\[- 3a = - 3b\]
\[\frac{a}{b} = 1.\]
\[\textbf{б)}\ \frac{a + 2b}{b + 2a} = - 3\]
\[a + 2b = - 3 \cdot (b + 2a)\]
\[a + 2b = - 3b - 6a\]
\[a + 6a = - 3b - 2b\]
\[7a = - 5b\]
\[\frac{a}{b} = - \frac{5}{7}.\]
\[\textbf{в)}\ \frac{99a + 8b}{4b - 100a} = 2\]
\[99a + 8b = 2 \cdot (4b - 100a)\]
\[99a + 8b = 8b - 200a\]
\[299a = 0b\]
\[\frac{a}{b} = 0.\ \]
