Страница 262 номер 1031, ГДЗ по алгебре за 9 класс к учебнику Мерзляка

1) \(a_{3} + a_{5} + a_{8} = 18\) и \(a_{2} + a_{4} = - 2\)
\(\left\{ \begin{aligned} & a_{1} + 2d + a_{1} + 4d + a_{1} + 7d = 18 \\ & a_{1} + d + a_{1} + 3d = - 2 \end{aligned} \right.\ \Rightarrow \begin{cases} 3a_{1} + 13d = 18 \\ 2a_{1} + 4d = - 2 \end{cases}\ \Rightarrow\)
\(\Rightarrow \begin{cases} 6a_{1} + 26d = 36 \\ - 6a_{1} - 12d = 6 \end{cases}\ \Rightarrow 14d = 42 \Rightarrow d = 3\)
\(2a_{1} + 4d = - 2 \Rightarrow a_{1} = \dfrac{- 2 - 4d}{2} = \dfrac{- 2 - 12}{2} = - 7\)
\(a_{1} = - 7;\ d = 3\)

2) \(a_{5} - a_{3} = - 4\) и \(a_{2}a_{4} = - 3\)
\(\begin{cases} a_{1} + 4d - a_{1} - 2d = - 4 \\ \left( a_{1} + d \right)\left( a_{1} + 3d \right) = - 3 \end{cases}\ \Rightarrow \begin{cases} 2d = - 4 \\ \left( a_{1} + d \right)\left( a_{1} + 3d \right) = - 3 \end{cases}\ \Rightarrow\)
\(\Rightarrow \begin{cases} d = - 2 \\ \left( a_{1} - 2 \right)\left( a_{1} - 6 \right) = - 3 \end{cases}\ \Rightarrow \begin{cases} d = - 2 \\ a_{1}^{2} - 8a_{1} + 12 = - 3 \end{cases}\)
\(a_{1}^{2} - 8a_{1} + 12 = - 3\)
\(a_{1}^{2} - 8a_{1} + 15 = 0\)
\(\left( a_{1} - 3 \right)\left( a_{1} - 5 \right) = 0\)
\(a_{1} = 3;\ d = - 2\)
\(a_{1} = 5;\ d = - 2\)

3) \(a_{2} + a_{4} + a_{6} = 36\) и \(a_{2}a_{3} = 54\)
\(\left\{ \begin{aligned} & a_{1} + d + a_{1} + 3d + a_{1} + 5d = 36 \\ & \left( a_{1} + d \right)\left( a_{1} + 2d \right) = 54 \end{aligned} \right.\ \Rightarrow \begin{cases} 3a_{1} + 9d = 36 \\ \left( a_{1} + d \right)\left( a_{1} + 2d \right) = 54 \end{cases}\ \Rightarrow\)
\(\Rightarrow \left\{ \begin{aligned} & a_{1} = 12 - 3d \\ & (12 - 3d + d)(12 - 3d + 2d) = 54 \end{aligned} \right.\)
\((12 - 3d + d)(12 - 3d + 2d) = 54\)
\((12 - 2d)(12 - d) = 54\)
\((6 - d)(12 - d) = 27\)
\(72 - 18d + d^{2} - 27 = 0\)
\(d^{2} - 18d + 45 = 0\)
\((d - 3)(d - 15) = 0\)
\(d = 3;\ a_{1} = 12 - 3 \cdot 3 = 3\)
\(d = 15;\ a_{1} = 12 - 3 \cdot 15 = - 33\)
\(a_{1} = 3;\ d = 3\)
\(a_{1} = - 33;\ d = 15\)

4) \(S_{5} - S_{2} - a_{5} = 0{,}1\) и \(a_{7} + S_{4} = 0{,}1;\)
\(\left\{ \begin{aligned} & S_{5} - S_{2} - a_{5} = 0{,}1 \\ & a_{7} + S_{4} = 0{,}1 \end{aligned} \right.\ \Rightarrow\)
\(\Rightarrow \left\{ \begin{aligned} & a_{1} + a_{2} + a_{3} + a_{4} + a_{5} - a_{1} - a_{2} - a_{5} = 0{,}1 \\ & a_{7} + \dfrac{a_{1} + a_{4}}{2} \cdot 4 = 0{,}1 \end{aligned} \right.\ \Rightarrow\)
\(\Rightarrow \left\{ \begin{aligned} & a_{3} + a_{4} = 0{,}1 \\ & a_{7} + 2a_{1} + 2a_{4} = 0{,}1 \end{aligned} \right.\ \Rightarrow\)
\(\Rightarrow \left\{ \begin{aligned} & a_{1} + 2d + a_{1} + 3d = 0{,}1 \\ & a_{1} + 6d + 2a_{1} + 2a_{1} + 6d = 0{,}1 \end{aligned} \right.\ \Rightarrow\)
\(\Rightarrow \begin{cases} 2a_{1} + 5d = 0{,}1 \\ 5a_{1} + 12d = 0{,}1 \end{cases}\ \Rightarrow \begin{cases} - 10a_{1} - 25d = - 0{,}5 \\ 10a_{1} + 24d = 0{,}2 \end{cases}\ \Rightarrow - d = - 0{,}3 \Rightarrow\)
\(\Rightarrow d = 0{,}3\)
\(2a_{1} + 5d = 0{,}1 \Rightarrow a_{1} = \dfrac{0{,}1 - 5d}{2} = \dfrac{0{,}1 - 5 \cdot 0{,}3}{2} = - 0{,}7\)
\(a_{1} = - 0{,}7;\ d = 0{,}3\)

5) \(S_{4} = 9\) и \(S_{6} = 22{,}5\).
\(\begin{cases} S_{4} = 9 \\ S_{6} = 22{,}5 \end{cases}\ \Rightarrow \begin{cases} \frac{{2a}_{1} + d \cdot 3}{2} \cdot 4 = 9 \\ \frac{2a_{1} + 5d}{2} \cdot 6 = 22{,}5 \end{cases}\ \Rightarrow \begin{cases} 2a_{1} + 3d = 4{,}5 \\ 2a_{1} + 5d = 7{,}5 \end{cases}\ \Rightarrow\)
\(\Rightarrow 2d = 3 \Rightarrow d = 1{,}5\)
\(2a_{1} + 3d = 4{,}5 \Rightarrow a_{1} = \dfrac{4{,}5 - 3 \cdot 1{,}5}{2} = 0\)
\(a_{1} = 0;\ d = 1{,}5\)

Ответ: 1) \(a_{1} = - 7;\ d = 3\); 2) \(a_{1} = 3;\ d = - 2\); \(a_{1} = 5;\ d = - 2\); 3) \(a_{1} = 3;\ d = 3\); \(a_{1} = - 33;\ d = 15\); 4) \(a_{1} = - 0{,}7;\ d = 0{,}3\); 5) \(a_{1} = 0;\ d = 1{,}5\).