ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 793

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\[\textbf{а)}\ 0,8x^{2} - 19,8x - 5 = 0\]

\[4x^{2} - 99 - 25 = 0\]

\[D = 9801 + 400 =\]

\[= 10\ 201 = 101^{2}\]

\[x_{1} = \frac{99 + 101}{8} = 25;\ \ \]

\[x_{2} = \frac{99 - 101}{8} = - \frac{1}{4}.\]

\[0,8x^{2} - 19,8x - 5 =\]

\[= \frac{4}{5} \cdot (x - 25)\left( x + \frac{1}{4} \right) =\]

\[= \left( \frac{1}{5}x - 5 \right)(4x + 1).\]

\[\textbf{б)}\ \ 3,5 - 3\frac{1}{3}x + \frac{2}{3}x^{2} = 0\]

\[\frac{2}{3}x^{2} - \frac{10}{3}x + \frac{7}{2} = 0\ \ \ \ \ | \cdot 6\]

\[4x^{2} - 20x + 21 = 0\]

\[D_{1} = 100 - 84 = 16\]

\[x_{1} = \frac{10 + 4}{4} = \frac{14}{4} = \frac{7}{2};\]

\[x_{2} = \frac{10 - 4}{4} = \frac{6}{4} = \frac{3}{2}.\]

\[3,5 - 3\frac{1}{3}x + \frac{2}{3}x^{2} =\]

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

\[= (x - 3,5)(x - 1).\]

\[\textbf{в)}\ x² + x\sqrt{2} - 2 = 0\]

\[D = \left( \sqrt{2} \right)^{2} + 4 \cdot 8 =\]

\[= 2 + 8 = 10\]

\[x_{1,2} = \frac{- \sqrt{2} \pm \sqrt{10}}{2};\]

\[x^{2} + x\sqrt{2} - 2 =\]

\[= \left( x + \frac{\sqrt{2} + \sqrt{10}}{2} \right) \cdot\]

\[\cdot \left( x + \frac{\sqrt{2} - \sqrt{10}}{2} \right).\]

\[\textbf{г)}\ x² - x\sqrt{6} + 1 = 0\]

\[D = \left( \sqrt{6} \right)^{2} - 4 \cdot 1 = 6 - 4 = 2\]

\[x_{1,2} = \frac{\sqrt{6} \pm \sqrt{2}}{2}\]

\[x^{2} - x\sqrt{6} + 1 =\]

\[= \left( x - \frac{\sqrt{6} + \sqrt{2}}{2} \right) \cdot\]

\[\cdot \left( x - \frac{\sqrt{6} - \sqrt{2}}{2} \right)\text{.\ }\]