Log3 (-cosx) - log9 (sinx) + 1/4 = -log9 (2) - вопрос №21162633391492 от egorhoyrovegiror 27.11.2021 06:34

Question

Алгебра: Log3 (-cosx) - log9 (sinx) + 1/4 = -log9 (2)

egorhoyrovegiror · Answer

4) 4+5(-3x+7)=-9
     4-15x+35+9=0
   -15x=-48
   x =3.2
5) -7+2(7x-2)=10
    -7+14x-4=10
    14x=21
   x =1.5
6) 10-2(-x-7)=9
    10+2x+14=9
     2x=-15
     x=-7.5
7) 8-4(-7x+8)=4
   8+28x-32=4
   28x=28
   x=1
8) -1+(7-6x)=2x+6
   -1+7-6x-2x=6
   -8x=0
   x=0
9) -2+3(1+2x)=-2x+3
   -2+3+6x+2x=3
   8x=2
   x=0.25
10) -3x+4=-10+5(-7-x)
   -3x+4=-10-35-5x
   2x=-49
   x=-24.5
11) 5x-4=4-3(5-2x)
   5x-4=4-15+6x
   x=-7
12) -1-4(-7+8x)=-2x-6
   -1+28-32x=-2x-6
   -30x=33
   x=-1.1
13) 4(7+x)-3x=7
   28+4x-3x=7
   x=-21