Log (3+x) числа (9-x^2) - 1/16log^2 (x+3) числа (x-3) ^2>=2

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Алгебра

Святослав

21 января, 02:33

Log (3+x) числа (9-x^2) - 1/16log^2 (x+3) числа (x-3) ^2>=2

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Артём · Answer 1

Log (x+3) (9-x^2) - (log (x+3) (x-3) ^2) ^2/16 > = 2 ОДЗ - 3 < x < 3

log (x+3) (x+3) + log (x+3) (3-x) - (log (x+3) (3-x) + log (x+3) (3-x)) ^2/16 > = 2

1+log (x+3) (3-x) - (log (x+3) (3-x) + log (x+3) (3-x)) ^2/16 > = 2

log (x+3) (3-x) - (log (x+3) (3-x) + log (x+3) (3-x)) ^2/16 > = 1

y=Log (x+3) (3-x)

y-2y^2/16 > = 1

4y-y^2 > = 4

y^2-4y+4 < = 0

y=2

log (x+3) (3-x) = 2

3-x = (x+3) ^2

x^2+7x+6=0

x1=-1 x2=-6

Ответ: х=-1