<span>1)2Cos</span>²<span>3</span>α <span>- Cos6</span>α = 2Cos²3α - 2Cos²α +1 = 1<span>
2) 2Cos</span>²α<span>/ 1+ Cos2</span>α= 2Cos²α/(1 + 2Cos²α -1) = 2Cos²α/2Cos²α = 1<span>
3) 1 - Cos (</span>π <span>- 2</span>α<span>)/ 1 - Sin</span>²α = 1+Cos2α/Сos²α= (1 +2Cos²α -1)/Cos²α =2<span>
4) Cos2 </span>α<span> + Sin</span>²α<span>/ 1 - Sin</span>²α= (Cos²α - Sin²α + Sin²α)/Сos²α = 1
График данного уравнения прилагается во вложении
5х-4y
x=0 y=13
5*0-4*13=-52
Ответ:-52
5x-4y
x=1,2 y=3,25
5*1,2-4*3,25=-7
Ответ:-7
Вроде правильно
1-(2*a+1)*(y+x)
2-(3*b+1)*(y-x)
5-(5*a-1)*(y+x)
6--(4*a-1)*(n-m)