Решебник по алгебре и начала математического анализа 10 класс Колягин Задание 336

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\[1)\ x^{2} + px + q = 0\]

\[По\ теореме\ Виета:\]

\[x_{1} + x_{2} = - p;\ \ \ x_{1} \cdot x_{2} = q.\]

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

\[x_{1}^{2} + x_{2}^{2} = p^{2} - 2q.\]

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

\[= p^{2} - 2q - 2q = p^{2} - 4q\]

\[x_{1} - x_{2} = \pm \sqrt{p^{2} - 4q};\]

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

\[= \pm \sqrt{p^{2} - 4q} \cdot ( - p) =\]

\[= \pm p\sqrt{p^{2} - 4q}.\]

\[2)\ x^{2} + px + q = 0\]

\[По\ теореме\ Виета:\]

\[x_{1} + x_{2} = - p;\ \ \ x_{1} \cdot x_{2} = q.\]

\[Из\ пункта\ 1):\]

\[x_{1}^{2} + x_{2}^{2} = p^{2} - 2q;\]

\[x_{1}^{2} - x_{2}^{2} = \pm p\sqrt{p^{2} - 4q}.\]

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

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

\[= - p \cdot \left( 2p^{2} - 2q - q \right) =\]

\[= - p \cdot \left( p^{2} - 3q \right) = 3pq - p^{3}.\]

\[x_{1}^{3} - x_{2}^{3} =\]

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

\[= \pm p\sqrt{p^{2} - 4q} \cdot\]

\[\cdot \left( p^{2} - 2q + q \right) =\]

\[= \pm p\sqrt{p^{2} - 4q} \cdot \left( p^{2} - q \right).\]