Question

$( \frac{1}{3} ) {}^{x {}^{2 } + 2x } > ( \frac{1}{9} {})^{16-x}$
Помогите решить​

Answer 1

$(\frac{1}{3})^{x^{2}+2x}>(\frac{1}{9})^{16-x}\\\\(\frac{1}{3})^{x^{2}+2x}>(\frac{1}{3})^{32-2x}\\\\0<\frac{1}{3}<1 \ \Rightarrow \ x^{2} +2x<32-2x\\\\x^{2} +2x-32+2x<0\\\\x^{2}+4x-32<0\\\\(x+8)(x-4)<0$

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x ∈ (- 8 ; 4)