Ответ:
-2y+x
Объяснение: -4x+y+5x-3y=x(-4+5)+y(1-3)= x(1)+y(-2)=x+(-2y)=x-2y
14000-100%
X-15%
14000*15/100=2100
14000+2100=16100
1)
(x+2)(2x-1)=2x^2-x+4x-2=2x^2+3x-2
(2-y)(y^2+3) =2y^2+6-y^3-3y
(а+4)(1-а)+а ^2=a-a^2+4-4a=-a^2+4-3 a
(b+2)(b^2-b+2)=b^3-b^2+2b+2b^2-2b^2+4=b^3-b^2+4
2) ab+2b+ac+2c=(a+c)^2
3) 2х(2-3х)(3х+2)=8х-18х^3.
2x(6x+4-9x^2-6x)= 8x-18x^3
2x(4-9x^2)=8x-18x^3
8x-18x^2=8x-18x^3
-18x^2+18x^3=8x-8x
-18x^2+18x^3=0
x^2(-18+18x)=0
x^2=0 -18+18x=0
18 x=18
x=1
4. Дано: ![\tt b_1=3x-5; \ \ b_2=2x; \ \ b_3=3x](https://tex.z-dn.net/?f=%5Ctt%20b_1%3D3x-5%3B%20%5C%20%5C%20b_2%3D2x%3B%20%5C%20%5C%20b_3%3D3x)
Найти: х
По свойству геометрической прогрессии:
![\displaystyle\tt {b_2}^2=b_1\cdot b_3\\ (2x)^2=(3x-5)\cdot3x\\4x^2=9x^2-15x\\5x^2-15x=0\\5x(x-3)=0\\1) \ 5x=0\\{} \ \ \ \ \ x=0 \ \ \ \O\\ 2) \ x-3=0\\{} \ \ \ \ x=3](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ctt%20%7Bb_2%7D%5E2%3Db_1%5Ccdot%20b_3%5C%5C%20%282x%29%5E2%3D%283x-5%29%5Ccdot3x%5C%5C4x%5E2%3D9x%5E2-15x%5C%5C5x%5E2-15x%3D0%5C%5C5x%28x-3%29%3D0%5C%5C1%29%20%5C%205x%3D0%5C%5C%7B%7D%20%5C%20%5C%20%5C%20%5C%20%5C%20x%3D0%20%5C%20%5C%20%5C%20%5CO%5C%5C%202%29%20%5C%20x-3%3D0%5C%5C%7B%7D%20%5C%20%5C%20%5C%20%5C%20x%3D3)
Ответ: х = 3.
5. Дано: b₁ = 2; b₃ - b₂ = 12
Найти: b₂; b₃
![\tt b_3-b_2=12\\b_1q^2-b_1q-12=0\\2q^2-2q-12=0 \ \ |:2\\q^2-q-6=0\\D=1+24=25=5^2\\q_1=\cfrac{1-5}{2}=-2 \ \ \ \O \\ q_2=\cfrac{1+5}{2}=3\\\\\\ b_2=b_1q=2\cdot3=6\\b_3=b_1q^2=2\cdot9=18](https://tex.z-dn.net/?f=%5Ctt%20b_3-b_2%3D12%5C%5Cb_1q%5E2-b_1q-12%3D0%5C%5C2q%5E2-2q-12%3D0%20%5C%20%5C%20%7C%3A2%5C%5Cq%5E2-q-6%3D0%5C%5CD%3D1%2B24%3D25%3D5%5E2%5C%5Cq_1%3D%5Ccfrac%7B1-5%7D%7B2%7D%3D-2%20%5C%20%5C%20%5C%20%5CO%20%5C%5C%20q_2%3D%5Ccfrac%7B1%2B5%7D%7B2%7D%3D3%5C%5C%5C%5C%5C%5C%20b_2%3Db_1q%3D2%5Ccdot3%3D6%5C%5Cb_3%3Db_1q%5E2%3D2%5Ccdot9%3D18)
Ответ: b₂ = 6; b₃ = 18.
6.
![\displaystyle\tt \left \{ {{b_1+b_4=252} \atop {b_2+b_3=60 \ }} \right. \ \Rightarrow \ \left \{ {{b_1+b_1q^3=252} \atop {b_1q+b_1q^2=60}}\ \Rightarrow \ \left \{ {{b_1(1+q^3)=252} \atop {b_1q(1+q)=60}}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ctt%20%5Cleft%20%5C%7B%20%7B%7Bb_1%2Bb_4%3D252%7D%20%5Catop%20%7Bb_2%2Bb_3%3D60%20%5C%20%7D%7D%20%5Cright.%20%5C%20%5CRightarrow%20%5C%20%5Cleft%20%5C%7B%20%7B%7Bb_1%2Bb_1q%5E3%3D252%7D%20%5Catop%20%7Bb_1q%2Bb_1q%5E2%3D60%7D%7D%5C%20%5CRightarrow%20%5C%20%5Cleft%20%5C%7B%20%7B%7Bb_1%281%2Bq%5E3%29%3D252%7D%20%5Catop%20%7Bb_1q%281%2Bq%29%3D60%7D%7D)
Из нижнего уравнения: ![\displaystyle\tt b_1=\frac{60}{q(1+q)}](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ctt%20b_1%3D%5Cfrac%7B60%7D%7Bq%281%2Bq%29%7D)
Подставим в верхнее:
![\displaystyle\tt \frac{60(1+q^3)}{q(1+q)}=252; \ \ q\neq0; \ q\neq -1\\\\\\ \frac{60(1+q)(1-q+q^2)}{q(1+q)}=252\\\\\\ \frac{60(1-q+q^2)}{q}=252\\\\ 60-60q+60q^2=252q\\\\60q^2-312q+60=0 \ \ |:12\\\\5q^2-26q+5=0\\\\ D=676-100=576=24^2\\\\ q_1=\frac{26-24}{2\cdot5}=0.2\\\\ q_2=\frac{26+24}{2\cdot5}=5](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Ctt%20%5Cfrac%7B60%281%2Bq%5E3%29%7D%7Bq%281%2Bq%29%7D%3D252%3B%20%5C%20%5C%20q%5Cneq0%3B%20%5C%20q%5Cneq%20-1%5C%5C%5C%5C%5C%5C%20%5Cfrac%7B60%281%2Bq%29%281-q%2Bq%5E2%29%7D%7Bq%281%2Bq%29%7D%3D252%5C%5C%5C%5C%5C%5C%20%5Cfrac%7B60%281-q%2Bq%5E2%29%7D%7Bq%7D%3D252%5C%5C%5C%5C%2060-60q%2B60q%5E2%3D252q%5C%5C%5C%5C60q%5E2-312q%2B60%3D0%20%5C%20%5C%20%7C%3A12%5C%5C%5C%5C5q%5E2-26q%2B5%3D0%5C%5C%5C%5C%20D%3D676-100%3D576%3D24%5E2%5C%5C%5C%5C%20q_1%3D%5Cfrac%7B26-24%7D%7B2%5Ccdot5%7D%3D0.2%5C%5C%5C%5C%20q_2%3D%5Cfrac%7B26%2B24%7D%7B2%5Ccdot5%7D%3D5)
Получаем две прогрессии:
убывающая (q=0.2)
![\tt b_1=\cfrac{60}{q(1+q)}=\cfrac{60}{0.2(1+0.2)}=250\\\\b_2=b_1q=250\cdot 0.2=50\\\\ b_3=b_2q=50\cdot 0.2=10\\\\ b_4=b_3q=10\cdot 0.2=2](https://tex.z-dn.net/?f=%5Ctt%20b_1%3D%5Ccfrac%7B60%7D%7Bq%281%2Bq%29%7D%3D%5Ccfrac%7B60%7D%7B0.2%281%2B0.2%29%7D%3D250%5C%5C%5C%5Cb_2%3Db_1q%3D250%5Ccdot%200.2%3D50%5C%5C%5C%5C%20b_3%3Db_2q%3D50%5Ccdot%200.2%3D10%5C%5C%5C%5C%20b_4%3Db_3q%3D10%5Ccdot%200.2%3D2)
возрастающая (q=5)
![\tt b_1=\cfrac{60}{q(1+q)}=\cfrac{60}{5(1+5)}=2\\\\b_2=b_1q=2\cdot 5=10\\\\ b_3=b_2q=10\cdot 5=50\\\\ b_4=b_3q=50\cdot 5=250](https://tex.z-dn.net/?f=%5Ctt%20b_1%3D%5Ccfrac%7B60%7D%7Bq%281%2Bq%29%7D%3D%5Ccfrac%7B60%7D%7B5%281%2B5%29%7D%3D2%5C%5C%5C%5Cb_2%3Db_1q%3D2%5Ccdot%205%3D10%5C%5C%5C%5C%20b_3%3Db_2q%3D10%5Ccdot%205%3D50%5C%5C%5C%5C%20b_4%3Db_3q%3D50%5Ccdot%205%3D250)