Question

решите пожалуйста , эти 5 неравенств

Answer 1

$9*3 leq frac{1}{9} * (frac{1}{27} )^{ frac{1}{x} } xneq0!\ 3^{2+1+2} leq 27^{- frac{1}{x}} \ 3^5 leq 3^{- frac{3}{x} } \ 3 textgreater 1 = textgreater 5 leq -frac{3}{x} \ frac{5x+3}{x} leq 0 \ -0.6 leq x < 0$

$7 leq ( frac{1}{49})^{frac{2}{x}}*(frac{1}{343})^{frac{1}{x^2}}\ 7^1 leq 7^{-frac{4}{x} -frac{3}{x^2}} \ 7 textgreater 1 = textgreater 1 leq - frac{4}{x} -frac{3}{x^2} \ frac{x^2+4x+3}{x^2} leq 0 \ -3 leq x leq -1$

$sqrt{32} *2^{-4x^2} geq 8^{3x} \ 2^{2.5-4x^2} geq 2^{9x} \ 2 textgreater 1 = textgreater 2.5-4x^2 geq 9x \ 8x^2 + 18x - 5 leq 0 \ x_{12} = frac{-9 б sqrt{81+40}}{8} = frac{-9бsqrt{121}}{8} = frac{-9 б 11}{8} \ x_1 = -2.5, x_2 = 0.25 \ -2.5 leq x leq 0.25$

$125*(frac{1}{5})^{x^2} leq (frac{1}{25})^{-4x}\ 5^{3-x^2} leq 5^{8x} \ 5 textgreater 1 = textgreater 3-x^2 leq 8x \ x^2+8x-3 geq 0\ x_{12} = frac{-4 б sqrt{16+3}}{1} = -4 б sqrt{19} \ x leq -4-sqrt{19}, x geq -4+sqrt{19}$

$log_{frac{1}{9}}(x+8)+log_{frac{1}{9}}x geq -1 , x textgreater 0!\ log_{frac{1}{9}}(x^2+8x) geq -1 \ -0.5log_3(x^2+8x) geq -1 \ log_3(x^2+8x) leq log_39\ 3 textgreater 1 = textgreater x^2+8x leq 9 \ x^2+8x-9 leq 0$
0 < x ≤ 1