(3-x)^2-x^2=15
(9-6x+x^2)-x^2=15
9-6x+x^2-x^2=15
9-6x=15
6x=-6
x=-1
Вроде так.
Решение
1) (x² - 7x + 12)/(x - 3) = 0
x² - 7x + 12 = 0, x₁ = 3 ; x₂ = 4
x - 3 ≠ 0, x ≠ 3
Ответ: х = 4
2) (x³ - 64x) / ( x + 4) = 0
x³ - 64x = 0
x + 4 ≠ 0, x ≠ - 4
x³ - 64x = 0
x(x² - 64) = 0
x₁ = 0
x² - 64 = 0
x² = √64
x₂ = - 8
x₃ = 8
Ответ: x₁ = 0 ; x₂ = - 8 ; x₃ = 8
2215, 2216, 2217, 2218, 2219, 2220, 2221
Вычислить двойной интеграл
![\int\limits^2_0 {} \, dy \int\limits^1_0 {(x^2+2y)} \, dx](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E2_0+%7B%7D+%5C%2C+dy+%5Cint%5Climits%5E1_0+%7B%28x%5E2%2B2y%29%7D+%5C%2C+dx++)
Решение:
Найдем внутренний интеграл:
![\int\limits^1_0 {(x^2+2y)} \, dx =( \frac{x^3}{3}+2yx) \left. \right|_0^1=\frac{1^3}{3}+2y*1-\frac{0^3}{3}-2y*0=\frac{1}{3}+2y](https://tex.z-dn.net/?f=%5Cint%5Climits%5E1_0+%7B%28x%5E2%2B2y%29%7D+%5C%2C+dx+%3D%28+%5Cfrac%7Bx%5E3%7D%7B3%7D%2B2yx%29++%5Cleft.+%5Cright%7C_0%5E1%3D%5Cfrac%7B1%5E3%7D%7B3%7D%2B2y%2A1-%5Cfrac%7B0%5E3%7D%7B3%7D-2y%2A0%3D%5Cfrac%7B1%7D%7B3%7D%2B2y)
Результат подставим во внешний интеграл
![\int\limits^2_0 { (\frac{1}{3}+2y) } \, dy =( \frac{1}{3}y+y^2)\left. \right|_0^2= \frac{2}{3}+2^2-\frac{0}{3}-0^2= \frac{2}{3}+4=4 \frac{2}{3}](https://tex.z-dn.net/?f=+%5Cint%5Climits%5E2_0+%7B+%28%5Cfrac%7B1%7D%7B3%7D%2B2y%29+%7D+%5C%2C+dy+%3D%28+%5Cfrac%7B1%7D%7B3%7Dy%2By%5E2%29%5Cleft.+%5Cright%7C_0%5E2%3D+%5Cfrac%7B2%7D%7B3%7D%2B2%5E2-%5Cfrac%7B0%7D%7B3%7D-0%5E2%3D+%5Cfrac%7B2%7D%7B3%7D%2B4%3D4+%5Cfrac%7B2%7D%7B3%7D++)
Ответ:
![4 \frac{2}{3}](https://tex.z-dn.net/?f=4+%5Cfrac%7B2%7D%7B3%7D+)