А)12а+2
Если а=4,05 , то 12*4,05+2=48,6+2=50,6
б)2.5b-(8*b+1.7)
если b=0,5 , то 2,5*0,5-(8*0,5+1,7)=1,25-(4+1,7)=1,25-5,7=-4,45
в)16,4-(7,2d+2.7)
Если d=0.25 , то 16,4-(7,2*0,25+2,7)=16,4-(1,8+2,7)=16,4-4,5=11,9
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)
(8у-1)^2-49у^2=0
64у^2-16у+1-49у^2=0
15у^2-16у+1=0
D=16^2-4*15*1=256-60=196
y1=-(-16)+14/2*15=1
y2=16-14/30=1/15
![\sqrt{x-4}](https://tex.z-dn.net/?f=+%5Csqrt%7Bx-4%7D+)
![\geq](https://tex.z-dn.net/?f=+%5Cgeq+)
22
Возводим левую и правую части в квадрат.
![( \sqrt{x-4} )^{2} \geq 22^{2}](https://tex.z-dn.net/?f=%28+%5Csqrt%7Bx-4%7D+%29%5E%7B2%7D++%5Cgeq+22%5E%7B2%7D+)
x-4≥484
x≥484+4
x≥488
Ответ: x≥488
А1=-3 d=5 a20=a1+19d=-3+19*5=-3+95=95-3=92
S20=(a1+a20)*20\2=(-3+92)*10=(92-3)*10=890