Решебник по алгебре 9 класс Мерзляк Задание 155

\[\boxed{\text{155\ (155).}\text{\ }\text{Еуроки\ -\ ДЗ\ без\ мороки}}\]

\[1)\ y = |x - 2|\]

\[y = \left\{ \begin{matrix} x - 2;\ \ \ если\ x \geq 2 \\ - x + 2;если\ x < 2 \\ \end{matrix} \right.\ \text{\ \ \ }\]

\[2)\ y = |x + 3| - 1\]

\[x \geq - 3:\]

\[y = x + 3 - 1 = x + 2\]

\[x < - 3:\]

\[y = - x - 3 - 1 = - x - 4\ \]

\[3)\ y = |x - 1| + x\]

\[x \geq 1:\]

\[y = x - 1 + x = 2x - 1\]

\[x < 1:\]

\[y = - x + 1 + x = 1\]

\(\ \)

\[\boxed{\mathbf{155.\ }Еуроки\ - \ ДЗ\ без\ мороки}\]

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

\[\frac{a + 1^{\backslash a + 1}}{a - 1} - \frac{a^{\backslash a - 1}}{a + 1} =\]

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

\[\frac{3a + 1}{a^{2} - 1} \cdot \frac{a^{2} + a}{3a + 1} =\]

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

\[Тождество\ доказано.\]