\[\boxed{\text{554\ (554).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]
\[1)\ \]
\[\textbf{а)}\ x^{2} - 5x + 6 = 0\] \[D = 25 - 24 = 1\] \[x_{1,2} = \frac{5 \pm \sqrt{1}}{2} = \frac{5 \pm 1}{2}\] \[x_{1} = \frac{6}{2} = 3;\ \ x_{2} = \frac{4}{2} = 2\] |
\[6x^{2} - 5x + 1 = 0\] \[D = 25 - 24 = 1\] \[x_{1,2} = \frac{5 \pm \sqrt{1}}{2 \cdot 6} = \frac{5 \pm 1}{12}\] \[x_{1} = \frac{4}{12} = \frac{1}{3};\ \ x_{2} = \frac{6}{12} = \frac{1}{2}\] |
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\[Корни\ уравнений - взаимно\ \] \[обратные\ числа.\] |
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\[\textbf{б)}\ 2x^{2} - 13x + 6 = 0\] \[D = 169 - 48 = 121\] \[x_{1,2} = \frac{13 \pm \sqrt{121}}{2 \cdot 2} = \frac{13 \pm 11}{4}\] \[x_{1} = \frac{2}{4} = \frac{1}{2};\ \ x_{2} = \frac{24}{4} = 6\] |
\[6x^{2} - 13x + 2 = 0\] \[D = 169 - 48 = 121\] \[x_{1,2} = \frac{13 \pm \sqrt{121}}{2 \cdot 6} = \frac{13 \pm 11}{12}\] \[x_{1} = \frac{2}{12} = \frac{1}{6};\ \ x_{2} = \frac{24}{12} = 2\] |
\[Корни\ уравнений - взаимно\ \] \[обратные\ числа.\] \[2)\ \] \[Результаты\ двух\ сравниваемых\ \] \[уравнений\ являются\ \] \[обратными\ числами.\] \[3)\ \] \[ax^{2} + bx + c = 0\] \[D = b^{2} - 4ac\] \[x_{1} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a} =\] \[= \frac{- b \pm \sqrt{D}}{2a}.\] \[cx^{2} + bx + a = 0\] \[D = b^{2} - 4ac\] \[x_{2} = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2c} =\] \[= \frac{- b \pm \sqrt{D}}{2c}.\] \[x_{1} \cdot x_{2} = \frac{- b - \sqrt{D}}{2a} \cdot \frac{- b + \sqrt{D}}{2c} =\] \[= \frac{\left( - b - \sqrt{D} \right)\left( - b + \sqrt{D} \right)}{2a \cdot 2c} =\] \[= \frac{b^{2} - D}{4ac} = \frac{b^{2} - \left( b^{2} - 4ac \right)}{4ac} =\] \[= \frac{b^{2} - b^{2} + 4ac}{4ac} = \frac{4ac}{4ac} = 1.\] \[Следовательно,\ корни\ \] \[уравнений\ ax^{2} + bx + c = 0\ и\ \] \[cx^{2} + bx + a = 0\ являются\ \] \[парами\ взаимно\ обратных\ \] \[чисел.\] \[Что\ и\ требовалось\ доказать.\] |