А) (2у² - 3)(у² + 2) = 2у^4 +4y² - 3y² -6 = 2y^4 + y² - 6;
б) (a² - b²)(a - b) = a³ - a²b - ab² + b³;
в) (m² + 3n)(m² - n) = m^4 - m²n +3m²n - 3n² = m^4 + 2m²n - 3n².
Y=4-x² y=2-x S-?
4-x²=2-x
x²-x-2=0 D=9 √D=3
x₁=-1 x₂=2
S=₋₁²(4-x²-2+x)dx=₋₁²(2+x-x²)dx=(2x+x²/2-x³/3) ₋₁|²=
=2*2+2²/2+2³/3-(2*(-1)+(-1)²/2-(-1)³/3)=4+2-8/3-(-2+1/2+1/3)=6-8/3-(-1,5+1/3)=
=6-8/3+1,5-1/3=7,5-3=4,5.
Ответ: S=4,5 кв. ед.
![|[(a+b)(a-b)]|=|[a,a]-[a,b]+[b,a]-[b,b]|=|\vec{o}-[a,b]-[a,b]-\vec{o}|=|(-2)*[a,b]|=|-2|*|[a,b]|=2*|a|*|b|*sin(\frac{\pi}{2})=2*3*4*1=24](https://tex.z-dn.net/?f=%7C%5B%28a%2Bb%29%28a-b%29%5D%7C%3D%7C%5Ba%2Ca%5D-%5Ba%2Cb%5D%2B%5Bb%2Ca%5D-%5Bb%2Cb%5D%7C%3D%7C%5Cvec%7Bo%7D-%5Ba%2Cb%5D-%5Ba%2Cb%5D-%5Cvec%7Bo%7D%7C%3D%7C%28-2%29%2A%5Ba%2Cb%5D%7C%3D%7C-2%7C%2A%7C%5Ba%2Cb%5D%7C%3D2%2A%7Ca%7C%2A%7Cb%7C%2Asin%28%5Cfrac%7B%5Cpi%7D%7B2%7D%29%3D2%2A3%2A4%2A1%3D24)
* ![[a,a]=\vec{o}](https://tex.z-dn.net/?f=%5Ba%2Ca%5D%3D%5Cvec%7Bo%7D)
и ![[b,b]=\vec{o}](https://tex.z-dn.net/?f=%5Bb%2Cb%5D%3D%5Cvec%7Bo%7D)
, так как вектора в каждом из этих двух векторных произведений, очевидно, коллинеарны.
* ![[b,a]=-[a,b]](https://tex.z-dn.net/?f=%5Bb%2Ca%5D%3D-%5Ba%2Cb%5D)
по свойству векторного произведения.
* 
