<em>1) (6sin(π/12))*(cos(π/12))=3sinπ/6=3*0.5=</em><em>1.5</em>
<em>2) sin²(5α)-cos²(5α)= </em><em>-cos10α</em>
<em>3)(2tg2α)/(1-tg²(2α))=</em><em>tg4α</em>
<em>4)sin( A/6)cos(A/6)=</em><em>0.5sin (A/3)</em>
<em>5) 20sin²(4α)cos²(4α)=(5*(2sin(4α))*cos(4α))²=</em><em>5sin²8α</em>
<em>6)(sin3α)(cos3α)=</em><em>0.5sin6α</em>
1.1)=3(a+2b)
2)=4(3m-4n)
3)=5c(2k-3p)
4)=8a(x+1)
5)=5b(1-5c)
6)=7x(2x+1)
7)=n^5(n^5-1)
8)=m^6(1+m)
2.1)=266256-516*513=1548
2)0,343+0,7*0,51=0,7
3)0,0016-0,0088*1,2=-0,00896
3. 1)D=<span><span><span><span>(<span>−1</span>)^</span>2</span>−<span><span>4·1</span>·0</span></span>=<span>1−0</span></span>=<span>1
x1=-(-1)+1/2*1=2/2=1
x2=-(-1)-1/2*1=0/2=0
2)D=</span><span><span><span>15^2</span>−<span><span>4·1</span>·0</span></span>=<span>225−0</span></span>=<span>225=15
x1=-15+15/2*1=0/2=0
x2=-15-15/2*1=-30/2=-15
3)D=</span><span><span><span><span><span>(<span>−30</span>)^</span>2</span>−<span><span>4·5</span>·0</span></span>=<span>900−0</span></span>=900=30
x1=-(-30)+30/2*5=60/10=6
x2=-(-30)-30/2*5=0/10=0
4)</span><span><span><span><span>18^2</span>−<span><span>4·14</span>·0</span></span>=<span>324−0</span></span>=324=18
x1=-18+18/2*14=0/28=0
x2=-18-18/2*14=-36/28=-1,28</span>
Объяснение:
а - длина прямоугольника;
b - ширина прямоугольника.
а=3+3b
|cosx|≤1. так что, чтобы найти минимум и максимум просто подставляем -1 и 1. 3*1+5=8 и -3*1+5=2
E(x)=[2;8]
Ответ: 2 и 8.
108^3-7^3=(108-7)*(108^2+7^2+108*7)=101*(108^2+7^2+108*7) что и доказыает утверждение.