A(a-3b)/3b*1/7(a-3b)=a/21b
Преобразуем выражение
(6,2аb² - 3a) + (b - 7.2b²a) = 6,2аb² - 3a + b - 7.2b²a = b -3a - ab²
при а = -2,5 и b = 2
2 - 3 · (-2.5) - (-2.5) · 2² = 19.5
Ответ: 19.5
Данное уравнение не имеет целых корней.
Используем метод Феррари:
уравнение вида
![(1)\ x^4+ax^3+bx^2+cx+d=0](https://tex.z-dn.net/?f=%281%29%5C%20x%5E4%2Bax%5E3%2Bbx%5E2%2Bcx%2Bd%3D0)
с помощью замены
приводим к виду:
![(2)\ y^4+p*y^2+qy+r=0](https://tex.z-dn.net/?f=%282%29%5C%20y%5E4%2Bp%2Ay%5E2%2Bqy%2Br%3D0)
где:
![p=b-\frac{3a^2}{8}\\q=\frac{a^3}{8}-\frac{a*b}{2}+c\\r=-\frac{3a^4}{256}+\frac{a^2b}{16}-\frac{a*c}{4}+d](https://tex.z-dn.net/?f=p%3Db-%5Cfrac%7B3a%5E2%7D%7B8%7D%5C%5Cq%3D%5Cfrac%7Ba%5E3%7D%7B8%7D-%5Cfrac%7Ba%2Ab%7D%7B2%7D%2Bc%5C%5Cr%3D-%5Cfrac%7B3a%5E4%7D%7B256%7D%2B%5Cfrac%7Ba%5E2b%7D%7B16%7D-%5Cfrac%7Ba%2Ac%7D%7B4%7D%2Bd)
добавим и вычтем из левой части уравнения 2 выражение
, где s - некоторое число:
![y^4+p*y^2+qy+r=y^2+py^2+2sy^2+qy+r+s^2-2sy^2-s^2=\\=y^4+2sy^2+s^2+y^2(p-2s)+qy+r-s^2=\\=(y^4+2s*y^2+s^2)+(p-2s)(y^2+\frac{2*qy}{2*(p-2s)})+r-s^2=\\=(y^2+s)^2+(p-2s)(y^2+2(\frac{qy}{2(p-2s)}+\frac{q^2}{4(p-2s)^2})-\frac{\frac{q^2}{4(p-2s)^2}}{p-2s}+r-s^2=\\=(y^2+s)^2+(p-2s)(y+\frac{q}{2(p-2s)})^2+r^2-s^2-\frac{q^2}{4(p-2s)}](https://tex.z-dn.net/?f=y%5E4%2Bp%2Ay%5E2%2Bqy%2Br%3Dy%5E2%2Bpy%5E2%2B2sy%5E2%2Bqy%2Br%2Bs%5E2-2sy%5E2-s%5E2%3D%5C%5C%3Dy%5E4%2B2sy%5E2%2Bs%5E2%2By%5E2%28p-2s%29%2Bqy%2Br-s%5E2%3D%5C%5C%3D%28y%5E4%2B2s%2Ay%5E2%2Bs%5E2%29%2B%28p-2s%29%28y%5E2%2B%5Cfrac%7B2%2Aqy%7D%7B2%2A%28p-2s%29%7D%29%2Br-s%5E2%3D%5C%5C%3D%28y%5E2%2Bs%29%5E2%2B%28p-2s%29%28y%5E2%2B2%28%5Cfrac%7Bqy%7D%7B2%28p-2s%29%7D%2B%5Cfrac%7Bq%5E2%7D%7B4%28p-2s%29%5E2%7D%29-%5Cfrac%7B%5Cfrac%7Bq%5E2%7D%7B4%28p-2s%29%5E2%7D%7D%7Bp-2s%7D%2Br-s%5E2%3D%5C%5C%3D%28y%5E2%2Bs%29%5E2%2B%28p-2s%29%28y%2B%5Cfrac%7Bq%7D%7B2%28p-2s%29%7D%29%5E2%2Br%5E2-s%5E2-%5Cfrac%7Bq%5E2%7D%7B4%28p-2s%29%7D)
получим:
![(3)\ (y^2+s)^2+(p-2s)(y+\frac{q}{2(p-2s)})^2+r^2-s^2-\frac{q^2}{4(p-2s)}=0](https://tex.z-dn.net/?f=%283%29%5C%20%28y%5E2%2Bs%29%5E2%2B%28p-2s%29%28y%2B%5Cfrac%7Bq%7D%7B2%28p-2s%29%7D%29%5E2%2Br%5E2-s%5E2-%5Cfrac%7Bq%5E2%7D%7B4%28p-2s%29%7D%3D0)
Пусть s - корень уравнения
![(4)\ r^2-s^2-\frac{q^2}{4(p-2s)}=0](https://tex.z-dn.net/?f=%284%29%5C%20r%5E2-s%5E2-%5Cfrac%7Bq%5E2%7D%7B4%28p-2s%29%7D%3D0)
Тогда уравнение 3 примет вид:
![(5)(y^2+s)^2+(p-2s)(y+\frac{q}{2(p-2s)})^2=0](https://tex.z-dn.net/?f=%285%29%28y%5E2%2Bs%29%5E2%2B%28p-2s%29%28y%2B%5Cfrac%7Bq%7D%7B2%28p-2s%29%7D%29%5E2%3D0)
Избавляемся в уравнении 4 от знаменателя:
![r(p-2s)-s^2(p-2s)-\frac{q^2}{4}=0](https://tex.z-dn.net/?f=r%28p-2s%29-s%5E2%28p-2s%29-%5Cfrac%7Bq%5E2%7D%7B4%7D%3D0)
Раскроем скобки и получим:
![(6)\ 2s^3-ps^2-2rs+rp-\frac{q^2}{4}=0](https://tex.z-dn.net/?f=%286%29%5C%202s%5E3-ps%5E2-2rs%2Brp-%5Cfrac%7Bq%5E2%7D%7B4%7D%3D0)
Уравнение 6 называется кубической резольвентой уравнения 4 степени.
Разложим уравнение 5 на множители:
![(y^2+s)^2+(p-2s)(y+\frac{q}{2(p-2s)})^2=0\\(y^2+s)^2-(2s-p)(y-\frac{q}{2(2s-p)})^2=0\\(y^2+s^2)^2-(y*\sqrt{2s-p}-\frac{q}{2\sqrt{2s-p}})^2=0\\(y^2-y\sqrt{2s-p}+\frac{q}{2\sqrt{2s-p}}+s)(y^2+y\sqrt{2s-p}-\frac{q}{2\sqrt{2s-p}}+s)=0](https://tex.z-dn.net/?f=%28y%5E2%2Bs%29%5E2%2B%28p-2s%29%28y%2B%5Cfrac%7Bq%7D%7B2%28p-2s%29%7D%29%5E2%3D0%5C%5C%28y%5E2%2Bs%29%5E2-%282s-p%29%28y-%5Cfrac%7Bq%7D%7B2%282s-p%29%7D%29%5E2%3D0%5C%5C%28y%5E2%2Bs%5E2%29%5E2-%28y%2A%5Csqrt%7B2s-p%7D-%5Cfrac%7Bq%7D%7B2%5Csqrt%7B2s-p%7D%7D%29%5E2%3D0%5C%5C%28y%5E2-y%5Csqrt%7B2s-p%7D%2B%5Cfrac%7Bq%7D%7B2%5Csqrt%7B2s-p%7D%7D%2Bs%29%28y%5E2%2By%5Csqrt%7B2s-p%7D-%5Cfrac%7Bq%7D%7B2%5Csqrt%7B2s-p%7D%7D%2Bs%29%3D0)
Получим два квадратных уравнения:
![(7)\ y^2-y\sqrt{2s-p}+\frac{q}{2\sqrt{2s-p}}+s=0\\(8)\ y^2+y\sqrt{2s-p}-\frac{q}{2\sqrt{2s-p}}+s=0](https://tex.z-dn.net/?f=%287%29%5C%20y%5E2-y%5Csqrt%7B2s-p%7D%2B%5Cfrac%7Bq%7D%7B2%5Csqrt%7B2s-p%7D%7D%2Bs%3D0%5C%5C%288%29%5C%20y%5E2%2By%5Csqrt%7B2s-p%7D-%5Cfrac%7Bq%7D%7B2%5Csqrt%7B2s-p%7D%7D%2Bs%3D0)
Применяем этот метод для решения уравнения
![x^4+4x-1=0](https://tex.z-dn.net/?f=x%5E4%2B4x-1%3D0)
Перепишем уравнение в полном виде:
![x^4+0x^3+0x^2+4x-1=0](https://tex.z-dn.net/?f=x%5E4%2B0x%5E3%2B0x%5E2%2B4x-1%3D0)
коэффиценты:
a=0
b=0
c=4
d=-1
определяем p,q и r:
![p=b-\frac{3a^2}{8}=0\\q=\frac{a^3}{8}-\frac{a*b}{2}+c=0-0+c=4\\r=-\frac{3a^4}{256}+\frac{a^2b}{16}-\frac{a*c}{4}+d=0+0-0+d=-1](https://tex.z-dn.net/?f=p%3Db-%5Cfrac%7B3a%5E2%7D%7B8%7D%3D0%5C%5Cq%3D%5Cfrac%7Ba%5E3%7D%7B8%7D-%5Cfrac%7Ba%2Ab%7D%7B2%7D%2Bc%3D0-0%2Bc%3D4%5C%5Cr%3D-%5Cfrac%7B3a%5E4%7D%7B256%7D%2B%5Cfrac%7Ba%5E2b%7D%7B16%7D-%5Cfrac%7Ba%2Ac%7D%7B4%7D%2Bd%3D0%2B0-0%2Bd%3D-1)
ищем s:
![2s^3-ps^2-2rs+rp-\frac{q^2}{4}=0\\2s^3+2s-4=0\\s^3+s-2=0\\s=1\\1+1-2=0\Rightarrow s=1](https://tex.z-dn.net/?f=2s%5E3-ps%5E2-2rs%2Brp-%5Cfrac%7Bq%5E2%7D%7B4%7D%3D0%5C%5C2s%5E3%2B2s-4%3D0%5C%5Cs%5E3%2Bs-2%3D0%5C%5Cs%3D1%5C%5C1%2B1-2%3D0%5CRightarrow%20s%3D1)
подставляем p,q,r и s в квадратные уравнения 7 и 8:
![y^2-y\sqrt{2s-p}+\frac{q}{2\sqrt{2s-p}}+s=0\\y^2-y\sqrt{2}+\frac{4}{2\sqrt{2}}+1=0\\y^2-y\sqrt{2}+\sqrt{2}+1=0\\D=2-4(\sqrt{2}+1)<0 \Rightarrow x\in \varnothing](https://tex.z-dn.net/?f=y%5E2-y%5Csqrt%7B2s-p%7D%2B%5Cfrac%7Bq%7D%7B2%5Csqrt%7B2s-p%7D%7D%2Bs%3D0%5C%5Cy%5E2-y%5Csqrt%7B2%7D%2B%5Cfrac%7B4%7D%7B2%5Csqrt%7B2%7D%7D%2B1%3D0%5C%5Cy%5E2-y%5Csqrt%7B2%7D%2B%5Csqrt%7B2%7D%2B1%3D0%5C%5CD%3D2-4%28%5Csqrt%7B2%7D%2B1%29%3C0%20%5CRightarrow%20x%5Cin%20%5Cvarnothing)
![y^2+y\sqrt{2s-p}-\frac{q}{2\sqrt{2s-p}}+s=0\\y^2+y\sqrt{2}-\sqrt{2}+1=0\\D=2-4(-\sqrt{2}+1)=2+4\sqrt{2}-4=4\sqrt{2}-2\\y_{1,2}=\frac{-\sqrt{2} \pm \sqrt{4\sqrt{2}-2}}{2}](https://tex.z-dn.net/?f=y%5E2%2By%5Csqrt%7B2s-p%7D-%5Cfrac%7Bq%7D%7B2%5Csqrt%7B2s-p%7D%7D%2Bs%3D0%5C%5Cy%5E2%2By%5Csqrt%7B2%7D-%5Csqrt%7B2%7D%2B1%3D0%5C%5CD%3D2-4%28-%5Csqrt%7B2%7D%2B1%29%3D2%2B4%5Csqrt%7B2%7D-4%3D4%5Csqrt%7B2%7D-2%5C%5Cy_%7B1%2C2%7D%3D%5Cfrac%7B-%5Csqrt%7B2%7D%20%5Cpm%20%5Csqrt%7B4%5Csqrt%7B2%7D-2%7D%7D%7B2%7D)
Теперь находим x:
![x=y-\frac{a}{4}=y-0=y](https://tex.z-dn.net/?f=x%3Dy-%5Cfrac%7Ba%7D%7B4%7D%3Dy-0%3Dy)
Ответ: ![\frac{-\sqrt{2} \pm \sqrt{4\sqrt{2}-2}}{2}](https://tex.z-dn.net/?f=%5Cfrac%7B-%5Csqrt%7B2%7D%20%5Cpm%20%5Csqrt%7B4%5Csqrt%7B2%7D-2%7D%7D%7B2%7D)
20:2=10 24 делится на 6 и 4 ответ: а= 6 b= 4 проверка: 6*4=24 (6+4)*2= 20