A) log(3) (√x + 1) = -1
ОДЗ √x + 1 > 0
√x > - 1
x > 1
log(3) (√x + 1) = log(3) 1/3
√x + 1 = 1/3
√x = -2/3
нет корней
б) lg (2x² + 3x) - lg (6x + 2) = 0
ОДЗ 2x² + 3x > 0
6x + 2 > 0
x > 0
lg((2x² + 3x)/(6x + 2) = lg 1
(2x² + 3x)/(6x + 2) = 1
2x² + 3x = 6x + 2
2x² - 3x - 2 = 0
D = 9 + 16 = 25
x1 = (3 + 5)/4 = 8/4 = 2
x2 = (3 - 5)/4 = -2/4 = -1/2 - лишний корень
в) log(3) x = 4 log(x) 3 - 3
log(3) x = 4/log(3) x - 3 | * log(3) x
ОДЗ log(3) x ≠ 0
x≠ 1 и x > 0
log²(3) x = 4 - 3log(3) x
log²(3) x + 3log(3) x - 4 = 0
log(3) x = t
t² + 3t - 4 =0
D = 9 + 16 = 25
t1 = (-3 + 5)/2 = 2/2 = 1
t2 = (-3 - 5)/2 = -8/2 = -4
log(3) x = 1
x1 = 3
log(3) x = -4
x2 = 1/81