![1)~ MK=\{-1-(-2);3-(-4)\}=\{1;7\}\\ PM=\{-2-4;-4-4\}=\{-6;-8\}\\ \\ 2)~ |MK|=\sqrt{1^2+7^2}=\sqrt{50}=5\sqrt{2}\\ |PM|=\sqrt{(-6)^2+(-8)^2}=10](https://tex.z-dn.net/?f=1%29~+MK%3D%5C%7B-1-%28-2%29%3B3-%28-4%29%5C%7D%3D%5C%7B1%3B7%5C%7D%5C%5C+PM%3D%5C%7B-2-4%3B-4-4%5C%7D%3D%5C%7B-6%3B-8%5C%7D%5C%5C+%5C%5C+2%29~+%7CMK%7C%3D%5Csqrt%7B1%5E2%2B7%5E2%7D%3D%5Csqrt%7B50%7D%3D5%5Csqrt%7B2%7D%5C%5C+%7CPM%7C%3D%5Csqrt%7B%28-6%29%5E2%2B%28-8%29%5E2%7D%3D10)
![EF=2MK-3KP=\{2\cdot1+3\cdot6;2\cdot7+3\cdot8\}=\{20;38\}](https://tex.z-dn.net/?f=EF%3D2MK-3KP%3D%5C%7B2%5Ccdot1%2B3%5Ccdot6%3B2%5Ccdot7%2B3%5Ccdot8%5C%7D%3D%5C%7B20%3B38%5C%7D)
4) Скалярное произведение двух векторов:
MK * PM = 1*(-6) + 7*(-8) = - 62
![\cos\angle (MK,PM)=\dfrac{MK\cdot PM}{|MK|\cdot |PM|}=\dfrac{-62}{5\sqrt{2}\cdot10}=-0.62\sqrt{2}](https://tex.z-dn.net/?f=%5Ccos%5Cangle+%28MK%2CPM%29%3D%5Cdfrac%7BMK%5Ccdot+PM%7D%7B%7CMK%7C%5Ccdot+%7CPM%7C%7D%3D%5Cdfrac%7B-62%7D%7B5%5Csqrt%7B2%7D%5Ccdot10%7D%3D-0.62%5Csqrt%7B2%7D)
x₀= -b/2a - абсцисса вершины параболы y=ax²+bx+c
y=5x^2-20x+3
a=5; b=-20
x₀= -b/2a =20/(2·5)=2
y₀=5·2²-20·2+3
y₀=-17
О т в е т. (2;-17)
Введём новую переменную t. Пусть t = x² - 2x - 5
t² - 2t = 3
t² - 2t - 3 = 0
Решаем по теореме, обратной теореме Виета
{t1 + t2 = 2
{t1 * t2 = -3
t1 = -1
t2 = 3
x² - 2x - 5 = -1, или x² - 2x - 5 = 3
1) x² - 2x - 5 = -1
x² - 2x - 4 = 0
Решаем через дискриминант
D = b² - 4ac = (-2)² - 4 * (-4) = 20
x1 = (-b - √D) / (2a) = (2 - √20) / 2 = (2 - 2√5) / 2 = 1 - √5
x2 = (-b +√D) / (2a) = (2 + √20) / 2 = 1 + √5
2) x² - 2x - 5 = 3
x² - 2x - 8 = 0
{x1 + x2 = 2
{x1 * x2 = -8
x1 = -2
x2 = 4
Ответ:
x1 = 1 - √5
x2 = 1 + √5
x3 = -2
x4 = 4
A9=a1+d(n-1)=12+8d=88
8d=76
d=7.5
Решение задания смотри на фотографии