Question

Решите неравенство
9^х-8*3^х+12>0

Answer 1

Ответ: x∈(-∞; log(3;2))U(1+log(3;2);+∞)

Объяснение:

$9^x-8*3^x+12 > 0\\(3^2)^x-8*3^x+12 > 0\\3^2^x-8*3^x+12 > 0\\3^x=t \\= > t^2-8t+12 > 0$

t1=6    t2=2  => t∈(-∞;2)U(6;+∞)

$= > 3^x < 2\\x < log(3;2)\\3^x > 6 \\x > log(3;6) =1+log(3;2)$

(В log 3 - основание)

=> x∈(-∞; log(3;2))U(1+log(3;2);+∞)

Answer 2

${9}^{x} - 8 \times 3 {}^{x} + 12 > 0 \\ (3 {}^{x} ) {}^{2} - 8 \times {3}^{x} + 12 > 0 \\ {3}^{x} = a \: , \: a > 0 \\ a {}^{2} - 8a + 12 > 0 \ \\ \\ po \: \: \: teoreme \: \: \: vieta \\ {x}^{2} + bx + c = 0\\ x_{1} + x_{2} = - b\\ x_{1} x_{2} = c \\ \\ {a}^{2} - 8a +1 2 = 0 \\ a_{1} + a_{2} = 8 \\ a_{1}a_{2} = 12 \\ a_{1} = 2 \\ a_{2} = 6 \\ \\ (a - 2)(a - 6) > 0 \\ + + + (2) - - - (6) + + + \\ a < 2 \\ a > 6 \\ \\ 3 {}^{x} < 2 \\ 3 {}^{x} > 6 \\ \\ {3}^{x} < {3}^{ log_{3}(2) } \\ {3}^{x} > {3}^{ log_{3}(6) } \\ \\ x < log_{3}(2) \\ x > log_{3}(6) \\ \\ x \: \epsilon \: ( - \infty ; \: log_{3}(2) )U( log_{3}(6) ; \: + \infty )$