Cos10a*cos8a + cos8a*cos6a =
= cos8a * (cos10a + cos6a) =
= cos8a * 2cos (10a+6a)/2* cos(10a-6a)/2 =
= cos8a * 2cos(16a)/2 * cos(4a)/2 =
= cos8a * 2cos8a * cos2a =
= 2cos(2a) * cos^2(8a)
За скобки можно вынести 4 и 12...
получим: 4(х²+(1/х²)) + 12(х+(1/х)) - 47 = 0
замена: х+(1/х) = t
тогда t² = x² + 2 + (1/x²)
4(t² - 2) + 12t - 47 = 0
4t² + 12t - 55 = 0
D=144+16*55=32²
t₁;₂ = (-12±32)/8 = (-3±8)/2
х+(1/х) = -11/2 или х+(1/х) = 5/2
2х² + 11х + 2 = 0 или 2х² - 5х + 2 = 0
D=121-16=105 D=25-16=3²
х₁;₂ = (-11±√105)/4
х₃;₄ = (5±3)/4 х₃ = 0.5 х₄ = 2
10x - 9 - x = 702
9x = 702 + 9
9x = 711
x = 711/9
x = 79
3 ответ-это точно!)там без вариантов!!!!!!!!!!!!!!!!!!