Question

Допоможіть будь ласочка з алгеброю даю 100 балів ​

Answer 1

3.

$\displaystyle\bf\\\left \{ {{x - y = 3} \atop { {x}^{2} - xy - 2 {y}^{2} = 7}} \right. \\ \displaystyle\bf\\\left \{ {{x = y + 3} \atop {(y + 3) {}^{2} - y(y + 3) - 2 {y}^{2} = 7 }} \right. \\ \\ {y}^{2} + 6y + 9 - {y}^{2} - 3y - 2 {y}^{2} = 7 \\ - 2 {y}^{2} + 3y + 2 = 0 \\ 2 {y}^{2} - 3y - 2 = 0 \\ a =2 \\ b = - 3\\ c = - 2 \\ D = {b}^{2} - 4ac = ( - 3) {}^{2} - 4 \times 2 \times ( - 2) =\\= 9 + 16 = 25\\ y_{1} = \frac{3 - 5}{2 \times 2} = - \frac{2}{4} = - 0.5 \\ y_{2} = \frac{3 + 5}{2 \times 2} = \frac{8}{4} = 2 \\ \\ x_{1} = - 0.5 + 3 = 2.5 \\ x_{2} = 2 + 3 = 5$

Ответ: ( 2,5 ; - 0,5 ) и ( 5 ; 2 )

4.

$\displaystyle\bf a_{n} = a_{1} + (n - 1)d \\ \displaystyle\bf \\ \left \{ {{a_{5} = - 0.8} \atop {a_{11} = - 5 }} \right. \\ \displaystyle\bf \\ \left \{ {{a_{1} + 4d = - 0.8 \: \: | \times ( - 1)} \atop {a _{1} + 10d = - 5}} \right. \\ \displaystyle\bf\\ + \left \{ {{ - a_{1} - 4d = 0.8} \atop {a _{1} + 10d = - 5}} \right. \\ \\ 10d - 4d = - 5 + 0.8 \\ 6d = - 4.2 \\ d = - 4.2 \div 6 \\ d = - 0.7 \\ \\ a_{1} + 4 \times ( - 0.7) = - 0.8 \\ a_{1} - 2.8 = - 0.8 \\ a_{1} = - 0.8 + 2.8 \\ a_{1} = 2 \\ \\ S_{n} = \frac{a_{1} + a_{n}}{2} \times n \\ a_{20} = a_{1} + 19d =2 + 19 \times ( - 0.7) = 2 - 13.3 =\\= - 11.3 \\ S_{20} = \frac{a_{1} + a_{20}}{2} \times 20 = (2 - 11.3) \times 10 = \\ = - 9.3 \times 10 = - 93$