ГДЗ по алгебре 9 класс Макарычев Задание 224

\[\boxed{\text{224\ (224).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

Пояснение.

Решение.

\[\textbf{а)}\ 0,8x^{2} - 19,8x - 5\]

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

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

\[D = 9801 + 400 = 10\ 201 = 101^{2}\]

\[x_{1} = \frac{99 + 101}{8} = \frac{200}{8} =\]

\[= \frac{100}{4} = 25;\]

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

\[= - \frac{1}{4} = - 0,25.\]

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

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

\[= (x - 25)(0,8x + 0,2).\]

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

\[\frac{2}{3}x^{2} - \frac{10}{3}x + 3,5 = 0\ \ \ \ \ | \cdot 6\]

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

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

\[x_{1} = \frac{10 + 4}{4} = \frac{14}{4} = \frac{7}{2} = 3,5;\]

\[x_{2} = \frac{10 - 4}{4} = \frac{6}{4} = \frac{3}{2} = 1,5.\]

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

\[= \frac{2}{3} \cdot (x - 3,5)\left( x - \frac{3}{2} \right) =\]

\[= (x - 3,5)\left( \frac{2}{3}x - 1 \right).\]

\[\textbf{в)}\ x^{2} + x\sqrt{2} - 2\]

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

\[D = \left( \sqrt{2} \right)^{2} + 4 \cdot 8 = 2 + 8 = 10\]

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

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

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

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

\[\textbf{г)}\ x^{2} - x\sqrt{6} + 1\]

\[x^{2} - x\sqrt{6} + 1 = 0\]

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

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

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

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

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