умоляю... люди... помогитеее Пожалуйста. заранее огромное спасибо.. Created by sav3107. algebra-ru

5.25ˣ -6*5ˣ +5=05²ˣ - 6*5ˣ +5=0y=5ˣy² -6y+5=0D=36-20=16y₁=(6-4)/2=1y₂=(6+4)/2=5При y=15ˣ=1x=0При y=55ˣ=5x=1Ответ: 0; 1.4.b₁=5n=1 b₁₊₁=3b₁ b₂=3*5=15n=2 b₂₊₁=3b₂ b₃=3*15=45n=3 b₃₊₁=3b₃ b₄=3*45=1355; 15; 45; 13515²=5*45225=225Да, это геометрическая прогрессия.S₄=b₁+b₂+b₃+b₄=5+15+45+135=2003.a₂=a₁+da₃=a₁+2da₄=a₁+3da₈=a₁+7d{a₁+2d+a₁+3d=7{a₁+d+a₁+7d=13{2a₁+5d=7{2a₁+8d=132a₁=7-5d7-5d+8d=133d=13-73d=6d=22a₁=7-5*22a₁=-3a₁= -1.5a₁₀=a₁+9d= -1.5+9*2=-1.5+18=16.5S₁₀=[(-1.5+16.5)/2] * 10=5*15=75Ответ: а) a₁=-1.5; d=2 б) S₁₀=752.a) [(1-cos²α)+sin²α+2sinαcosα]/(sinα+cosα)==[sin²α+sin²α+2sinαcosα]/(sinα+cosα)==[2sin²α+2sinαcosα]/(sinα+cosα)==[2sinα(sinα+cosα)]/(sinα+cosα)=2sinαб) [sinФ/(1-cosФ)] - [cosФ/sinФ]==[sin²Ф-cosФ+cos²Ф] / [sinФ(1-cosФ)]==(1-cosФ)/[sinФ(1-cosФ)]=1/sinФ Реклама Реклама Новые вопросы в Алгебра Розв'язати систему нерівностей: 6^3×4^3×x^3:x^2=24^2 7xy - x2 - 5xy3 + Зух - 7xy + 2x2. 99×965+12-45×3-1+37+9-45=сколько будет? решите 64 и 66 пожалуйста !!!!!!!! (7x)^11*(49x)^2*7/(7x^2)^3*(343x)^4=56 Докажите неравенство: x ^ 2+6x+11 > 0 Подробно пожалуйста Дана функция f(x)=6x2. Вычисли f(−1)= розв'яжіть нерівність |х-2|+|х+1|≥6 x ^ 2+6x+11 > 0 помогите пожалуйста Предыдущий Следующий Вместе мы знаем больше Этот сайт использует cookies Политика Cookies . Вы можете указать условия хранения и доступ к cookies в своем браузере. 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умоляю... люди... помогитеее Пожалуйста. заранее огромное спасибо...

sav3107 @sav3107

October 2021 1 1 Report

умоляю... люди... помогитеее Пожалуйста. заранее огромное спасибо.

Answers & Comments

m11m

Verified answer

5.
25ˣ -6*5ˣ +5=0
5²ˣ - 6*5ˣ +5=0

y=5ˣ
y² -6y+5=0
D=36-20=16
y₁=(6-4)/2=1
y₂=(6+4)/2=5

При y=1
5ˣ=1
x=0

При y=5
5ˣ=5
x=1
Ответ: 0; 1.

4.
b₁=5
n=1 b₁₊₁=3b₁ b₂=3*5=15
n=2 b₂₊₁=3b₂ b₃=3*15=45
n=3 b₃₊₁=3b₃ b₄=3*45=135

5; 15; 45; 135

15²=5*45
225=225
Да, это геометрическая прогрессия.
S₄=b₁+b₂+b₃+b₄=5+15+45+135=200

3.
a₂=a₁+d
a₃=a₁+2d
a₄=a₁+3d
a₈=a₁+7d

{a₁+2d+a₁+3d=7
{a₁+d+a₁+7d=13

{2a₁+5d=7
{2a₁+8d=13

2a₁=7-5d
7-5d+8d=13
3d=13-7
3d=6
d=2

2a₁=7-5*2
2a₁=-3
a₁= -1.5

a₁₀=a₁+9d= -1.5+9*2=-1.5+18=16.5

S₁₀=[(-1.5+16.5)/2] * 10=5*15=75

Ответ: а) a₁=-1.5; d=2
б) S₁₀=75

2.
a) [(1-cos²α)+sin²α+2sinαcosα]/(sinα+cosα)=
=[sin²α+sin²α+2sinαcosα]/(sinα+cosα)=
=[2sin²α+2sinαcosα]/(sinα+cosα)=
=[2sinα(sinα+cosα)]/(sinα+cosα)=2sinα

б) [sinФ/(1-cosФ)] - [cosФ/sinФ]=
=[sin²Ф-cosФ+cos²Ф] / [sinФ(1-cosФ)]=
=(1-cosФ)/[sinФ(1-cosФ)]=1/sinФ