Вопрос:

Для каждого значения а решите уравнение: (x^2-(a+1)x+a)/(x-2)=0.

Ответ:

\[x^{2} - (a + 1)x + a = 0\]

\[D = (a + 1)^{2} - 4 \cdot 1 \cdot a =\]

\[= a^{2} + 2a + 1 - 4a =\]

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

\[x_{1,2} = \frac{a + 1 \pm \sqrt{(a - 1)^{2}}}{2} =\]

\[= \frac{a + 1 \pm (a - 1)}{2}.\]

\[1.\ \ При\ D = 0:\]

\[(a - 1)^{2} = 0\ \ \]

\[a - 1 = 0\ \ \]

\[a = 1.\]

\[x = \frac{1 + 1 \pm 0}{2} = \frac{2}{2} = 1.\]

\[2.\ При\ a \neq 1;a \neq 2:\]

\[x_{1} = \frac{a + 1 + a - 1}{2} = \frac{2a}{2} = a;\]

\[x_{2} = \frac{a + 1 - a + 1}{2} = \frac{2}{2} = 1.\]

\[3.\ При\ a = 2 \Longrightarrow x = 1.\]

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