Раскроем скобки:
x²-2x+7x-14-(x²+3x+4x+12)+2=0
x²-2x+7x-14-x²-3x-4x-12+2=0
-2x-24=0
-2x=24
x=-12
1) 13а - 2bc + 19 bc = 13 a + 17bc
2) 0.7b^2 + 20a + 2b^2= 2.7b^2+20a
3) 9.3c +4.5d^3 - 5.1d^3 = 9.3c - 0.7d^3
4) 10nm + 9x - 20nm = 9x - 10nm
5) 5xy - 34xy + 3.3a = 3.3a - 29xy
6) 0.8t^4 + 2.4c - 2.1t^4 = 2.4c - 1.3t^4
^ - знак степени
<h3>А) cos(π/19) + сos(3π/19) + cos(5π/19) + ... + cos(17π/19) = 1/2</h3>
Пусть φ = π/19, тогда f(φ) = cos(φ) + cos(3φ) + cos(5φ) + ... + cos(17φ) = 1/2
Домножим обе части на 2sin(φ) и применим известную формулу:
2cos(α)•sin(β) = sin(α + β) - sin(α - β)
f(φ)•2sin(φ) = 2cos(φ)•sin(φ) + 2cos(3φ)•sin(φ) + 2cos(5φ)•sin(φ) + ... + 2cos(15φ)•sin(φ) + 2cos(17φ)•sin(φ) = sin(2φ) + sin(4φ) - sin(2φ) + sin(6φ) - sin(4φ) + ... + sin(16φ) - sin(14φ) + sin(18φ) - sin(16φ) = sin(18φ)
f(φ)•2sin(φ) = sin(18φ) ; sin(18φ) = sin(18π/19) = sin(π - (π/19)) = sin(π/19)
f(φ)•2sin(φ) = sin(φ) ⇒ f(φ) = 1/2, ч.т.д.
<h3>Б) cos(6π/7) + cos(4π/7) + sin(11π/14) = - 1/2</h3>
sin(11π/14) = sin((π/2) + (2π/7)) = cos(2π/7) ⇒ cos(2π/7) + cos(4π/7) + cos(6π/7) = - 1/2
Пусть φ = π/7, тогда f(φ) = cos(2φ) + cos(4φ) + cos(6φ) = - 1/2
Таким ж образом, домножим обе части на 2sin(φ) и применим ту ж формулу:
f(φ)•2sin(φ) = 2cos(2φ)•sin(φ) + 2cos(4φ)•sin(φ) + 2cos(6φ)•sin(φ) = sin(3φ) - sin(φ) + sin(5φ) - sin(3φ) + sin(7φ) - sin(5φ) = sin(7φ) - sin(φ) = sin(7π/7) - sin(φ) = - sin(φ)
f(φ)•2sin(φ) = - sin(φ) ⇒ f(φ) = - 1/2 , ч.т.д.
