2(1-сos2 x)-5сos x+1>0
2сos2 x+5сos x-3<0
пусть sin x=t
2t2+5t-3<0
D=49. t1=0.5 t2=-3-посторонний корень
t =0,5
сos x=0.5
x=+_п/3+2пк
Ответ:(-п/3+2пк;п/3+2пк)
Q>0 a₁-a₂=8 a₃=4 a₅-?
a₁-a₁q=8 a₁(1-q)=8
a₁q²=4 a₁q=4
Разделим первое уравнение на второе:
(1-q)/q²=2
1-q=2q²
2q²+q-1=0 D=9
q₁=1/2 q₂=-1 ∉
a₁*(1/2)²=4
a₁/4=4 |×4
a₁=16. ⇒
a₅=a₁q⁴=16*(1/2)⁴=2⁴/2⁴=1.
Ответ: a₅=1.
1)х^2у+ху^2-2х-2у=ху(х+у)-2(х+у)=(ху-2)(х+у)
2)а^3+27=а^3+3^3=(а+3)(а^2-3а+9)
![C^{x+1}_{x+2}= \frac{(x+2)!}{((x+2)-(x+1))!(x+1)!}=\frac{(x+2)!}{(1)!(x+1)!}=\frac{(x+1)!(x+2)}{(x+1)!} =x+2](https://tex.z-dn.net/?f=C%5E%7Bx%2B1%7D_%7Bx%2B2%7D%3D+%5Cfrac%7B%28x%2B2%29%21%7D%7B%28%28x%2B2%29-%28x%2B1%29%29%21%28x%2B1%29%21%7D%3D%5Cfrac%7B%28x%2B2%29%21%7D%7B%281%29%21%28x%2B1%29%21%7D%3D%5Cfrac%7B%28x%2B1%29%21%28x%2B2%29%7D%7B%28x%2B1%29%21%7D+%3Dx%2B2)
![A^3_{x+2}= \frac{(x+2)!}{(x+2-3)!}= \frac{(x+2)!}{(x-1)!}=\frac{(x-1)!x(x+1)(x+2)}{(x-1)!}=x(x+1)(x+2)](https://tex.z-dn.net/?f=A%5E3_%7Bx%2B2%7D%3D+%5Cfrac%7B%28x%2B2%29%21%7D%7B%28x%2B2-3%29%21%7D%3D+%5Cfrac%7B%28x%2B2%29%21%7D%7B%28x-1%29%21%7D%3D%5Cfrac%7B%28x-1%29%21x%28x%2B1%29%28x%2B2%29%7D%7B%28x-1%29%21%7D%3Dx%28x%2B1%29%28x%2B2%29)
получаем
2(x+2)=x(x+1)(x+2)
x=-2 отбрасываем
2=x(x+1)
x²+x-2=0
D=1²+4*2=9
√D=3
x₁=(-1-3)/2=-2 отбрасываем
x₂=(-1+3)/2=1
Ответ x=1