1) x-16<span><0
x</span><span><16
2) 28x+4</span><span>>0
28x</span><span>>-4
x</span><span>>-1/7
3) 31x+63</span><span>>36x+23
31x-36x</span><span>>23-63
-5x</span><span>>-40
x</span><span>>8</span>
2sin x * cos x - sinx + cos x=-1
1+2sinxcosx - sinx+cosx=0
sin²x+cos²x-2sinxcosx + 4sinxcosx - sinx+cosx=0
(sinx - cos x)²+4sin x cos x-(sinx-cosx)=0
Пусть sinx - cos x = t, сделаем условие что t ∈ [-√2;√2]
Возведем оба части до квадрата
(sin x- cos x)²=t²
1-2sinxcosx=t²
2sinxcosx=1-t²
В результате замены переменных, получаем
t²+2(1-t²)-t=0
t²+2-2t²-t=0
-t²-t+2=0 |*(-1)
t²+t-2=0
D=b²-4ac=9; √D=3
t1=[-1+3]/2=1
t2=[-1-3]/2=-2 - ∉ [-√2;√2]
Сделаем обратную замену
sinx - cosx = 1
√2sin(x-π/4)=1
sin(x-π/4)=1/√2
![x- \frac{\pi}{4} =(-1)^k*\frac{\pi}{4}+\pi k,k \in Z \\ x=(-1)^k*\frac{\pi}{4}+\frac{\pi}{4}+\pi k,k \in Z](https://tex.z-dn.net/?f=x-+%5Cfrac%7B%5Cpi%7D%7B4%7D+%3D%28-1%29%5Ek%2A%5Cfrac%7B%5Cpi%7D%7B4%7D%2B%5Cpi+k%2Ck+%5Cin+Z+%5C%5C+x%3D%28-1%29%5Ek%2A%5Cfrac%7B%5Cpi%7D%7B4%7D%2B%5Cfrac%7B%5Cpi%7D%7B4%7D%2B%5Cpi+k%2Ck+%5Cin+Z)
2sinx cos x - sinx - cos x =1
-1+2sinxcosx-(sinx+cosx)=0
-(sin²x+cos²x+2sinxcosx) +4sinxcosx - (sinx+cosx)=0
-(sinx+cosx)²+4sin xcosx-(sinx + cosx)=0
пусть sinx+cosx =t ///// t∈ [-√2;√2]
Возведем оба части до квадрата
(sinx+cosx)²=t²
1+2sinxcosx=t²
2sinxcosx=t²-1
Получаем
-t²+2(t²-1)-t=0
-t²+2t²-2-t=0
t²-t-2=0
D=b²-4ac=1+8=9
t1=[1+3]/2=2 ∉ [-√2;√2]
t2=[1-3]/2=-1
Замена
sin x+ cos x=-1
√2sin(x+π/4)=-1
sin(x+π/4) = -1/√2
![x+ \frac{\pi}{4} =(-1)^{k+1}*\frac{\pi}{4}+\pi k,k \in Z \\ x=(-1)^{k+1}*\frac{\pi}{4}-\frac{\pi}{4}+\pi k,k \in Z](https://tex.z-dn.net/?f=x%2B+%5Cfrac%7B%5Cpi%7D%7B4%7D+%3D%28-1%29%5E%7Bk%2B1%7D%2A%5Cfrac%7B%5Cpi%7D%7B4%7D%2B%5Cpi+k%2Ck+%5Cin+Z+%5C%5C+x%3D%28-1%29%5E%7Bk%2B1%7D%2A%5Cfrac%7B%5Cpi%7D%7B4%7D-%5Cfrac%7B%5Cpi%7D%7B4%7D%2B%5Cpi+k%2Ck+%5Cin+Z)
<span>10(ax-1)=2a-5x-9
</span>10ax-10=2a-5x-9
10ax+5х=2a-9+10
(10a+5)х=2a+1
Уравнение вида сх=d имеет бесконечно много решений при с=d=0
10a+5=0
а=-0,5
2a+1
а=-0,5
<em><u>Ответ: -0,5</u></em>