<span> √54-14√5*(√5+7) = <span> </span><span>√54-70-98<span> </span><span>√5=3<span> </span><span>√6+28<span> </span><span>√5
Вроде так)) </span></span></span></span></span>
<span>Cos4xcos2x+sin4xsin2x>0,5
cos(4x-2x)>0,5
cos2x>0,5
-</span>π/3+2πk<2x<π/3+2πk
-π/6+πk<x<π/6+πk,k∈z
Ответ x∈(-π/6+πk;π/6+πk,k∈z)
121x^3+22x^2+x=0
х (121х^2 + 22х + 1)= 0
х= 0
121х^2 + 22х + 1= 0
Д= 22^2 - 4×121×1= 484 - 484= 0
х= -22/(2×121)= -0.09
Ответ: х= 0; х= -0.09.
√2≈ 1,414 √3≈1,732 π≈3,142
√2+√3≈1,414+1,732≈3,146
√2+√3<π