пошаговое объяснение:
a) √2cos2x< < 1 b)2㏒²₅x-10㏒₂₅x+2=0
cos2x< < 1/√2=√2/2 2log²₅x-10log₅^²x+2=0
2x=π/4+2πk, k∈z 2log²₅x-10/2log₅x+2=0
x=π/8+πk, k∈z пусть log₅x=t
2t²-5t+2=0
d=(-5)²-4*2*2=25-16=9
x₁=(5+3)/2*2=8/4=2
x₂=(5-3)/2*2=2/4=1/2=0.5
решил как уравнение.
х+ 4 4/19 = 6 2/19
x = 6 2/19 - 4 4/19
x = 5 21/19 - 4 4/19
x = 1 17/19
2)
25 - х = 8 3/14
x = 25 - 8 3/14
x = 24 14/14 - 8 3/14
x = 16 11/14
3)
32 - х = 9 18/35
x = 32 - 9 18/35
x = 31 35/35 - 9 18/35
x = 22 17/35