Решить систему уравнений {6x+3=8x-3(2y-4) 2(2y-3y)-4x=2y-8

6x + 3 = 8x - 6y + 12

2x - 6y + 8 = 0

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- 2y - 4x - 2y + 8 = 0

- 4y - 4x + 8 = 0

-x - y - 2 = 0

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2x - 6y + 8 = 0

- x - y - 2 = 0

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x - 7y + 6 = 0

x = 7y - 7

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-2y -4*(7y - 7) - 2y + 8 = 0

- 4y - 28y + 28 + 8 = 0

- 32y = - 36

y = 1.125

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x = 7y - 7 = 7.875 - 7 = 0.875

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

ответ:

cos(x)+cos(2x)+1=0

2cos^2(x)+cos(x)+0

w=cos(x)

это уравнение вида:

a*w^2+b*w+c=0

w1=-   √(d   )-b/2a

w2=√(d   )-b/2a

где d=b^2-4*a*c

d=1

w1=0

w2=-1/2

делаем обратную замену

cos(x)=w

x=πn+acos(w)

x=πn+acos(w)- π

x1= πn+acos(w1)

x1= πn+acos(0)

x1= πn+π/2

x2= πn+acos(w2)

x2= πn+acos(-1/2)

x2= πn+2π/3

x3= πn+acos(w1)- π

x3= πn-π+acos(0)

x3= πn-π/2

x4= πn+acos(w2)- π

x4= πn-π+acos(-1/2)

x4= πn-π/3

ответ  

x1= -π/2

x2= π/2

x3= i(log(2)-log(-1-√3 i))

x4= i(log(2)-log(-1+√3 i))

объяснение:

1/n*(n+1) = 1/n - 1/(n+1) используем эту формулу

1/(x + 2019)(x + 2020) + 1/(x + 2020)(x + 2021) + 1/(x + 2021)(x + 2022) + 1/(x + 2022)(x + 2023) = 1/

1/(x + 2019) - 1/(x + 2020) + 1/(x + 2020) - 1/(x + 2021) + 1/(x + 2021) - 1/(x + 2022) + 1/(x + 2022) - 1/(x + 2023) = 1/

1/(x + 2019) - 1/(x + 2023) = 1/

(x + 2023 - x - 2019)* = (x + 2019)(x + 2023)

4* = x² + 4042x + 2019*2023

x² + 4042x + 2019*2023 - 4* = 0

4* = 4*1 - 4 = 36

2019*2023 = (2021 - 2)(2021 + 2) = 4084441 - 4 = 4084437

x² + 4042 x + 84441 = 0

d = b² - 4ac = 4042² - 4*84441 = 4*2021² - 4*84441) = 4*(4084441 - 84441) = 4*4 = 2²*2000² = 4000²

x12 = (-4042 +- 4000)/2 = -4021   и -21

ответ -21 и -4021