(x² -3x -3)(x² +3x +1) =0
x² -3x -3 =0
D =9+12 =21
x1 =(3 -√21)/2
x2 =(3+√21)/2
x² +3x +1 =0
D =9 -4 =5
x3 =(-3 -√5)/2
x4 =(-3 +√5)/2
Решение
<span>Найдите угол между векторами а(– 1; – 1) и в(2; 0) .
cos(a</span>∧b) = (x₁ *x₂ + y₁ * y₂) / [IaI*IbI]
IaI = √[(-1)² + (-1)²] = √2
IbI = √(2² + 0²) = √4 = 2
cos(a∧b) = [(-1)*2 + (-1)*0] / (2√2) = - 2/(2√2) = - 1/√2
(a<span>∧b) = 3</span>π/4
<span>)4a^3 b-6a^2 b^2=2a^2 b(2a-3b)
2)5x^2 y+10 xy^2=5xy(x+2y)
3) 14m^3 n-21 m^2n^3=7m^2n(2m-3n^2)
4)5x^3-15 x^2 y+ 20xy^2=x(5x^2-xy+20y^2)
5)2a^2y-6ay^2+8y=2y(a^2-3ay+4)
6)6ax-9a^2+15ax^2=3a(2x-3a+5x^2)</span>