Question

Решите неравенства:

a) ㏒₇ (2 - x) ≤ ㏒₇ (2x² - x)

б) ㏒₁/₂(x² - 1) < - 3

в) lg(7⁶⁻²ˣ+3) - lg39 > lg4 - 1g3

г) ㏒₂ₓ₊₁⁽⁵⁻²ˣ⁾ >1

Answer 1

N1

log 7 (2 - x) =< log 7 (2x^2 - x)

2 - x =< 2x^2 - x

2x^2 - 2 >= 0

x € (-беск. ; -1] U [1 ; +беск.)

N2

log 0,5 (x^2 - 1) < -3

log 0,5 (x^2 - 1) < log 0,5 (8)

x^2 - 1 > 8

x^2 - 9 > 0

x € (-беск. ; -3) U (3 ; +беск.)

N3

lg (7^(6 - 2x) + 3) - lg (39) > lg (4) - lg (3)

lg (7^(6 - 2x) + 3) > lg (39) + lg (4) - lg (3)

lg (7^(6 - 2x) + 3) > lg (52)

7^(6x - 2) + 3 > 52

7^(6x - 2) > 49

6x - 2 > 2

6x > 4

x > 2/3

N4

log 2x + 1 (5 - 2x) > 1

log 2x + 1 (5 - 2x) > log 2x + 1 (2x + 1)

5 - 2x > 2x + 1

- 4x > - 4

x < 1

log 2x + 1 (5 - 2x) > 1

log 2x + 1 (5 - 2x) > log 2x + 1 (2x + 1)

5 - 2x < 2x + 1

-4x < -4

x > 1

2x + 1 > 0

x € (-1/2 ; 0)

5 - 2x > 0

x € (0 ; 5/2)

{x € (-1/2 ; 0) x - не существует

{x > 1

{x € (0 ; 5/2) x € (0 ; 1)

{x < 1

Ответ : (0 ; 1