ГДЗ по алгебре 8 класс Макарычев ФГОС Задание 795

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\[mx^{2} + (m - 3)x - 3 = 0\]

\[D = (m - 3)^{2} + 4 \cdot 3 \cdot m =\]

\[= m^{2} - 6m + 9 + 12m =\]

\[= m^{2} + 6m + 9 =\]

\[= (m + 3)^{2}\]

\[x_{1} = \frac{- m + 3 + m + 3}{2m} =\]

\[= \frac{6}{2m} = \frac{3}{m};\]

\[x_{2} = \frac{- m + 3 - m - 3}{2m} =\]

\[= \frac{- 2m}{2m} = - 1.\]

\[при\ \ m = - 3:\ \ \]

\[x_{1} = - 1;\]

\[m = - 1:\ \ \]

\[x_{1} = - 3;\]

\[m = 3:\ \ \]

\[x_{1} = 1;\]

\[m = 1:\ \ \]

\[x_{1} = 3.\]