Log3 (5-4x)= log3 (x-1) log3 (7-x)> 1 log1/3 (3-5x)< -3 решить неравенство?

Log₃ (5  - 4x) = log₃ (x - 1) одз: 5 - 4x > 0, x < 5/4 x - 1 > 0, x > 1  одз: 1   < x <   5/4, 1 < x < 1,25  5 - 4x = x - 1 - 4x - x = - 1 - 5 -5x = - 6 x = 6/5 x = 1,2 2) log₃ (7 - x) > 1 одз: 7 -  x > 0, x < 7 3 > 1 7 - x > 3 - x > 3 - 7 -x > - 4 x < 4   x  ∈ (-  ≈ ; 4) 3) log₁/₃ (3 - 5x) < - 3 0 < 1/3 < 1 3 - 5x > (1/3)⁻³ -5x > 3³ - 3 - 5x > 27 - 3 - 5x > 24 x < - 4,8

3х-9у=3(х-3у)

16z-20y=4(4z-5y)

ab-bc=b(a-c)

4cx-acx=cx(4-a)

ad+bd+cd=d(a+b+c)

abx-acx-adx=ax(b-c-d)

а) x= 6+ 2y

      3(6+2y)+2y=-6

      18 + 6y+2y =-6

8y=-24

y=-3

x=0

 

б) x-y=0

2x+3y=-5

x=y

2y+3y=-5

5y=-5

y=-1

x=-1