Question

Помогите решить неравенство

Answer 1

$1)quad 25cdot 0,04^{2x} textgreater 0,2^{x(3-x)}\\25cdot ( frac{1}{25} )^{2x} textgreater (frac{1}{5})^{3x-x^2}\\5^{2-4x} textgreater 5^{x^2-3x}\\2-4x textgreater x^2-3x\\x^2+x-2 textless 0\\(x-1)(x+2) textless 0qquad +++(-2)---(1)+++\\xin (-2,1)$

$2)quad 4^{x}-2^{x} textless 12\\2^{2x}-2^{x}-12 textless 0\\t=2^{x} textgreater 0; ,; t^2-t-12 textless 0; ,; ; t_1=-3,; t_2=4\\(t+3)(t-4) textless 0qquad +++(-3)---(4)+++\\-3 textless t textless 4; ; ; to quad 0 textless 2^{x} textless 4\\2^{x} textless 2^2\\x textless 2\\xin (-infty ,2)$

$3)quad ( frac{1}{4})^{x}-32cdot (frac{1}{8})^{x^2-1} textless 0\\2^{-2x}-2^5cdot 2^{-3(x^2-1)} textless 0\\2^{-2x}-2^{-3x^2+8} textless 0\\2^{-2x} textless 2^{-3x^2+8}\\-2x textless -3x^2+8\\3x^2-2x-8 textless 0\\D/4=25,; ; x_1=-frac{4}{3}; ,; ; x_2=2\\3(x+frac{4}{3})(x-2) textless 0qquad +++(- frac{4}{3} )---(2)+++\\xin (- frac{4}{3} ,2)$

$4)quad 3cdot 9^{x}+11cdot 3^{x} textless 4\\3cdot 3^{2x}+11cdot 3^{x}-4 textless 0\\t=3^{x} textgreater 0; ,; ; 3t^2+11t-4 textless 0\\D=169; ,; ; t_1=-4,; ; t_2=frac{1}{3}\\3(t+4)(t-frac{1}{3}) textless 0qquad +++(-4)---(frac{1}{3})+++\\-4 textless t textless frac{1}{3}quad to quad 0 textless 3^{x} textless frac{1}{3}\\3^{x} textless 3^{-1}\\x textless -1\\xin (-infty ,-1)$