Нужно воспользоваться правилом Лопиталя
<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
<span>х-2у=3</span>
x=4-y
.
(4-y)-2y=3
-8y+2y^2=3
2y^2+2y-3=0
Через D решишь
<span>5√3=√(5^2*3)=√75
2а√3а=√((2a)^2*3a)=√(4a^2*3a)= √(12a^3)</span><span>
0.2√10=√0.2^2*10=√0.4</span>
15376-12996=2380..........