Решение на фото. Успехов тебе!
1. y=log₂(2x+3)
![y'= \frac{1}{(2x+3)ln2} *2= \frac{2}{(2x+3)ln2}](https://tex.z-dn.net/?f=y%27%3D%20%5Cfrac%7B1%7D%7B%282x%2B3%29ln2%7D%20%2A2%3D%20%5Cfrac%7B2%7D%7B%282x%2B3%29ln2%7D)
2.y=1/3cos(3x-π/2)-π³-e², x₀=π/3
y'=1/3 (-sin(3x-π/2))*3=-sin(3x-π/2)
y'(x₀)=-sin(3*π/3-π/2)=-sin(π-π/2)=-sin(π/2)=-1
3. y=x², x₀=0,25
y'=2x
k=y'(x₀)=2*0,25=0,5
4. sin(π/6-3x)-1/2=0
sin(π/6-3x)=1/2
π/6-3x=π/2+2πn
3x=π/6-π/2+2πn=π/6-3π/6+2πn=-2π/6+2πn=-π/3+2πn
x=-π/9+2πn/3 , n∈Z
2
a)(5-3x)/(x-3)(x+3)+(7x-1)/(2x(x+3)=(10x-6x²+7x²-21x-x+3)/[2x(x²-9)]=
=(x²-12x+3)/[2x(x²-9)]
b)1/2(x-y)²-(x-y-2)/4(x-y)²=(2-x+y+2)/4(x-y)²=(4-x+y)/4(x-y)²
3
a)7/a-4/(a-2b)-(a-b)/(a-2b)(a+2b)=(7a²-28b²-4a²-8ab-a²+ab)/[a(a²-4b²)]=
=(2a²-7ab-28b²)/[a(a²-4b²)]
b)(a²-a)/(a+1)(a²-a+1)-1/(1+a)=(a²-a-a²+a-1)/(a³+1)=-1/9a³+1)
4
a)(a+2)/2(a-2)-(a-2)/3(a+2)-(a²+4)/6(a-2)(a+2)=
=(3a²+6a+6a+12-2a²+4a+4a-8-a²-4)/6(a²-4)=20a/6(a²-4)=10a/3(a²-4)
a=-4
-40/3*(16-4)=-40/36=-10/9=-1 1/9
b)b/(a(a-b)+a/b(a+b)-(a²+b²)/a(a-b)(a+b)=(ab²+b³+a³-a²b-a²b-b³)/ab(a²-b²)=
=(ab²+a³-2a²b)/ab(a²-b²)=a(a²-2ab+b²)/ab(a²-b²)=(a-b)²/b(a-b)(a+b)=
(a-b)/b(a+b)
a=6;b=-1
(6+1)/[-1*(6-1)]=7/(-5)=-1,4
5
(a-2)/a(a+2)+(a+2)/a(a-2)-(2a²+8)/[a(a-2)(a+2)]=
=(a²-4a+4+a²+4a+4-2a²-8)/[a(a-2)(a+2)]=0
Углом равнобокой трапеция ABCD <ABH = 33 градусам
Пары смежных углов:
∠AOC ∠COB
∠COB ∠BOD
∠BOD ∠DOA
∠DOA ∠AOC
Т.е. 4 пары смежных углов
Вертикальные углы
∠DOA ∠COB
∠AOC ∠DOB
2 пары вертикальных углов