Question

Решить неравенство (С3) для 10-классника
$4 logx_{} 4-3log 4x_{} 4-4log frac{x}{16} 4 geq 0$

Answer 1

$4log_{x}4-3log_{4x}4-4log_{frac{x}{16}}4 geq 0; ,; ; ODZ:; x textgreater 0, ; xne 1\\log_{a}b=frac{1}{log_{b}a}\\frac{4}{log_4x}-frac{3}{log_44x}-frac{4}{log_4frac{x}{16}} geq 0\\frac{4}{log_4x}-frac{3}{log_44+log_4x}-frac{4}{log_4x-log_44^2} geq 0\\frac{4}{log_4x}-frac{3}{1+log_4x}-frac{4}{log_4x-2} geq 0\\t=log_4x,; ; frac{4}{t}-frac{3}{1+t}-frac{4}{t-2} geq 0\\frac{4(1+t)(t-2)-3t(t-2)-4t(1+t)}{t(1+t)(t-2)} geq 0\\frac{4t^2-4t-8-3t^2+6t-4t-4t^2}{t(t+1)(t-2)} geq 0$

$frac{-3t^2-2t-8}{t(t+1)(t-2)} geq 0\\frac{3t^2+2t+8}{t(t+1)(t-2)} leq 0\\3t^2+2t+8=0,; ; D=4-4cdot 3cdot 8 textless 0; to ; 3t^3+2t+8 textgreater 0; pri ; xin R\\arightarrow ; ; t(t+1)(t-2) textless 0\\---(-1)+++(0)---(2)+++\\tin (-infty,-1)U(0,2)\\ left { {{log_4x textless -1} atop {0 textless log_4x textless 2}} right. ; left { {{log_4x textless log_44^{-1}} atop {log_41 textless log_4x textless log_44^2}} right. ; left { {{x textless frac{1}{4}} atop {1 textless x textless 16}} right. ,; ; x textgreater 0,; xne 1\\xin (0,frac{1}{4})U(1,16)$