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Уравнение \(x^2 + 7x + 8 = 0\) имеет корни \(x\_1\) и \(x\_2\) , вычисли.

\(\dfrac{1}{x\_1} + \dfrac{1}{x\_2}\) .

Так как \(x\_1 + x\_2 = -7\) , а \(x\_1 \cdot x\_2 = 8\) , то \(\dfrac{1}{x\_1} + \dfrac{1}{x\_2} = \dfrac{x\_1 + x\_2}{x\_1 \cdot x\_2} = \dfrac{-7}{8}\) .

\((x\_1 - x\_2)^2\) .

\((x\_1 - x\_2)^2 = (x\_1 + x\_2)^2 - 4 x\_1 \cdot x\_2 = (-7)^2 - 4 \cdot 8 = 49 - 32 = 17\) .

а) \(x^2\_1 + x^2\_2 = x^2\_1 + 2x\_1x\_2 + x^2\_2 - 2x\_1x\_2 = \) [ ];

б) \(x^3\_1 + x^3\_2 = \) [ ];

в) \(\dfrac{1}{x^2\_1} + \dfrac{1}{x^2\_2} =\) [ ].