Решить неравенство (С3) для 10-классника [tex]4 logx_{} 4-3log 4x_{} 4-4log \frac{x}{16}

Ответы 1

$4log_{x}4-3log_{4x}4-4log_{\frac{x}{16}}4 \geq 0\; ,\; \; ODZ:\; x\ \textgreater \ 0, \; xe 1\\\\log_{a}b=\frac{1}{log_{b}a}\\\\\frac{4}{log_4x}-\frac{3}{log_44x}-\frac{4}{log_4\frac{x}{16}} \geq 0\\\\\frac{4}{log_4x}-\frac{3}{log_44+log_4x}-\frac{4}{log_4x-log_44^2} \geq 0\\\\\frac{4}{log_4x}-\frac{3}{1+log_4x}-\frac{4}{log_4x-2} \geq 0\\\\t=log_4x,\; \; \frac{4}{t}-\frac{3}{1+t}-\frac{4}{t-2} \geq 0\\\\\frac{4(1+t)(t-2)-3t(t-2)-4t(1+t)}{t(1+t)(t-2)} \geq 0\\\\\frac{4t^2-4t-8-3t^2+6t-4t-4t^2}{t(t+1)(t-2)} \geq 0$ $\frac{-3t^2-2t-8}{t(t+1)(t-2)} \geq 0\\\\\frac{3t^2+2t+8}{t(t+1)(t-2)} \leq 0\\\\3t^2+2t+8=0,\; \; D=4-4\cdot 3\cdot 8\ \textless \ 0\; \to \; 3t^3+2t+8\ \textgreater \ 0\; pri \; x\in R\\\\\arightarrow \; \; t(t+1)(t-2)\ \textless \ 0\\\\---(-1)+++(0)---(2)+++\\\\t\in (-\infty,-1)U(0,2)\\\\ \left \{ {{log_4x\ \textless \ -1} \atop {0\ \textless \ log_4x\ \textless \ 2}} ight. \; \left \{ {{log_4x\ \textless \ log_44^{-1}} \atop {log_41\ \textless \ log_4x\ \textless \ log_44^2}} ight. \; \left \{ {{x\ \textless \ \frac{1}{4}} \atop {1\ \textless \ x\ \textless \ 16}} ight. ,\; \; x\ \textgreater \ 0,\; xe 1\\\\x\in (0,\frac{1}{4})U(1,16)$
- Автор:
  soniamullins
- 5 лет назад
- 0
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