<span><span>(x+2)^4-4(x+2)^2-5=0
Замена
y^4-4y^2-5=0
y^2=5
y^2=-1-нет решений
(x+2)^2=5
x+2=+-</span>√5
x=√5-2
x=-√5-2
</span>
Найдем cosa:
![cosa=б\sqrt{1-sin^2a}=б\sqrt{1-16/25}=б\sqrt{9/25}=б3/5](https://tex.z-dn.net/?f=cosa%3D%D0%B1%5Csqrt%7B1-sin%5E2a%7D%3D%D0%B1%5Csqrt%7B1-16%2F25%7D%3D%D0%B1%5Csqrt%7B9%2F25%7D%3D%D0%B13%2F5)
но т.к. a во 2 четв. и сos во 2 четв. отриц то
cosa=-3/5
Вычеслим:
cos(pi/6+a)=cospi/6*cosa-sinpi/6*sina=√3/2 *cosa-1/2 *sina=√3/2 *(-3/5) - 1/2 *4/5=-3√3/10-2/5
1)(a^2-5a)+(b^2-5b)=a(a-5)+b(a-5)=(a-5)(a+b)=(6.6-5)(6.6+0.4)=1.5*7=10.5
2)a^2-ab-2a+2b=a(a-2)-b(a-2)=(a-2)(a-d)=(7/20-2)(7/20-0.15)=-33/20*20/100=0.33