Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 128

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\[2x^{2} - 3x - 2 = 0\ \ \ \ \ \ |\ :2\]

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

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

\[1)\ \frac{1^{\backslash x_{2}}}{x_{1}} + \frac{1^{\backslash x_{1}}}{x_{2}} = \frac{x_{2} + x_{1}}{x_{1} \cdot x_{2}} =\]

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

\[2)\ x_{1}^{2} + x_{2}^{2} = x_{1}^{2} + 2x_{1}x_{2} + x_{2}^{2} -\]

\[- 2x_{1}x_{2} = \left( x_{1} + x_{2} \right)^{2} - 2x_{1}x_{2} =\]

\[= \left( \frac{3}{2} \right)^{2} - 2 \cdot ( - 1) = \frac{9}{4} + 2 =\]

\[= 2\frac{1}{4} + 2 = 4\frac{1}{4} = 4,25.\]

\[3)\ x_{1}^{3} + x_{2}^{3} =\]

\[= \left( x_{1} + x_{2} \right)\left( x_{1}^{2} - x_{1}x_{2} + x_{2}^{2} \right) =\]

\[= \frac{3}{2} \cdot (4,25 + 1) = \frac{3}{2} \cdot 5,25 =\]

\[= \frac{3}{2} \cdot 5\frac{1}{4} = \frac{3}{2} \cdot \frac{21}{4} = \frac{63}{8} = 7,875.\]