1) sin²β - cos²(α - β) + 2cosα·cosβ·cos(α - β) = sin²β + cos(α - β)·(2cosα·cosβ - cos(α - β)) = sin²β + cos(α - β)·(2cosα·cosβ - (cosα·cosβ + sinα·sinβ)) = sin²β + (cosα·cosβ + sinα·sinβ)·(cosα·cosβ - sinα·sinβ) = sin²β + cos²α·cos²β - sin²α·sin²β = sin²β·(1 - sin²α) + cos²α·cos²β = sin²β·cos²α + cos²α·cos²β = cos²α·(sin²β + cos²β) = cos²α
2) cos²β + cos²(α - β) - 2cosα·cosβ·cos(α - β) = cos²β + cos(α - β)·(cos(α - β) - 2cosα·cosβ) = cos²β + cos(α - β)·(cosα·cosβ + sinα·sinβ - 2cosα·cosβ) = cos²β + (cosα·cosβ + sinα·sinβ)·(sinα·sinβ - cosα·cosβ) = cos²β + sin²α·sin²β - cos²α·cos²β = cos²β·(1 - cos²α) + sin²α·sin²β = cos²β·sin²α + sin²α·sin²β = sin²α·(sin²β + cos²β) = sin²α
X=0.125 вот ответ пожалуйста
( (2a)/(2a+b) - 4aˇ2/(4aˇ2+4ab+bˇ2)) : ( 2a/(4aˇ2-bˇ2) + 1/(b-2a))=
=(2a(2a+b)-4aˇ2)/(2a+b)ˇ2 : (2a-2a-b)/(4aˇ2-bˇ2)=
=(4aˇ2+2ab-4aˇ2)/(2a+b)ˇ2 . (2a+b)(2a-b)/(-b)=
=2ab/(2a+b)(2a+b) .(2a+b)(2a-b)/(-b)=-2a(2a-b)/(2a+b)=(-4aˇ2+2ab)/(2a+b)
(1/(x+1) - 3/(xˇ3+1) + 3/(xˇ2-x+1)) . (x - (2x-1)/(x+1)=
=(xˇ2-x+1-3+3x+3)/(xˇ2-x+1)(x+1) . (xˇ2+x-2x-1)/(x+1)=
=(xˇ2+2x+1)/(xˇ2-x+1)(x+1) . (xˇ2-x+1)/(x+1)=
=(x+1)(x+1)/(x+1)(x+1)=1
Решение смотри на фотографии
<span>(8y</span>²<span>+4y)(5y</span>³<span>+2y</span>²<span>+7y</span>⁴<span>) =
= 40у</span>⁵+20у⁴+16у⁴+8у³+56у⁶+28у⁵ =
= 56у⁶+68у⁵+36у⁴+8у³