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Точки пересечения парабол:
![y=x^2-4x+3\; ,\; \; y=-x^2+6x-5\\\\x^2-4x+3=-x^2+6x-5\\\\2x^2-10x+8=0\\\\x^2-5x+8=0\; \; \to \; \; x_1=1,\; x_2=4\; (teorema\; Vieta)\\\\S= \int\limits^4_1 {(-x^2+6x-5-(x^2-4x+3))} \, dx = \int\limits^4_1 {(-2x^2+10x-8)} \, dx =\\\\=(-\frac{2x^3}{3}+5x^2-8x)|_1^4=-\frac{128}{3}+80-32-(-\frac{2}{3}+5-8)=\\\\=\frac{-126}{3}+51=-42+51=9](https://tex.z-dn.net/?f=y%3Dx%5E2-4x%2B3%5C%3B%20%2C%5C%3B%20%5C%3B%20y%3D-x%5E2%2B6x-5%5C%5C%5C%5Cx%5E2-4x%2B3%3D-x%5E2%2B6x-5%5C%5C%5C%5C2x%5E2-10x%2B8%3D0%5C%5C%5C%5Cx%5E2-5x%2B8%3D0%5C%3B%20%5C%3B%20%5Cto%20%5C%3B%20%5C%3B%20x_1%3D1%2C%5C%3B%20x_2%3D4%5C%3B%20%28teorema%5C%3B%20Vieta%29%5C%5C%5C%5CS%3D%20%5Cint%5Climits%5E4_1%20%7B%28-x%5E2%2B6x-5-%28x%5E2-4x%2B3%29%29%7D%20%5C%2C%20dx%20%3D%20%5Cint%5Climits%5E4_1%20%7B%28-2x%5E2%2B10x-8%29%7D%20%5C%2C%20dx%20%3D%5C%5C%5C%5C%3D%28-%5Cfrac%7B2x%5E3%7D%7B3%7D%2B5x%5E2-8x%29%7C_1%5E4%3D-%5Cfrac%7B128%7D%7B3%7D%2B80-32-%28-%5Cfrac%7B2%7D%7B3%7D%2B5-8%29%3D%5C%5C%5C%5C%3D%5Cfrac%7B-126%7D%7B3%7D%2B51%3D-42%2B51%3D9)
2 ^ ( X + 5 ) = 16
2 ^ ( X + 5 ) = 2 ^ 4
X + 5 = 4
X = - 1
(x-2)(-x-1) = 0
-x^2-x+2x+2 = 0
-x^2+x+2 = 0 |*(-1)
x^2-x-2 = 0
D = (-1)^2-4*1*(-2) = 1+8 = 9.
x1 = (1+\/9)/2 = (1+3)/2 = 4/2 = 2.
x2 = (1-\/9)/2 = (1-3)/2 = -2/2 = -1.
Ответ : -1; 2.
1) x(3+x) <=0
Ответ: x принадлежит [ 0 ; -3]
2) 5 > x^2
![\sqrt{5}](https://tex.z-dn.net/?f=+%5Csqrt%7B5%7D+)
> x > -
![\sqrt{5}](https://tex.z-dn.net/?f=+%5Csqrt%7B5%7D+)