Вопрос:

Упростите выражение: (4a/(a^2-3a+2)+2/(a^2-1))∶(2a+4)/(a^2-a-2)-a/(a-1).

Ответ:

\[= \frac{4a^{2} + 6a - 4}{2 \cdot (a - 1)(a + 2)} - \frac{a}{a - 1} =\]

\[= \frac{2 \cdot \left( 2a^{2} + 3a - 2 \right)}{2 \cdot (a - 1)(a + 2)} - \frac{a}{a - 1} =\]

\[= \frac{2a - 1 - a}{a - 1} = \frac{a - 1}{a - 1} = 1\]

\[Дополнительные\ вычисления:\]

\[a^{2} - 3a + 2 = 0\]

\[a_{1} + a_{2} = 3\]

\[a_{1} \cdot a_{2} = 2\]

\[\Longrightarrow a_{1} = 2;\ \ a_{2} = 1.\]

\[a^{2} - 3a + 2 = (a - 2)(a - 1).\]

\[a^{2} - a - 2 = 0\]

\[a_{1} + a_{2} = 1\]

\[a_{1} \cdot a_{2} = - 2\]

\[\Longrightarrow a_{1} = 2;\ a_{2} = - 1.\]

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

\[2a^{2} + 3a - 2 = 0\]

\[a_{1} + a_{2} = - 1,5\]

\[a_{1} \cdot a_{2} = - 1\]

\[\Longrightarrow a_{1} = - 2;\ \ a_{2} = \frac{1}{2}.\]

\[2a^{2} + 3a - 2 =\]

\[= 2 \cdot (a + 2)\left( a - \frac{1}{2} \right) =\]

\[= (a + 2)(2a - 1).\]

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