Задача 55276 13. sqrt(7-log2x^2) + log2x^4 >...
Условие
sqrt(7-log2x^2) + log2x^4 > 4
14.
lg^2 ((x+2)^2(x+5)) / 5 < lg^2 (x+5)/20
Решение
[red]ОДЗ[/red]: x^2 >0 ⇒[red] x ≠ 0[/red]
Пусть [m]t=log_{2}x^2[/m], тогда [m]log_{2}x^4=log_{2}(x^2)^2=2log_{2}x^2=2t[/m]
Неравенство принимает вид:
[m]\sqrt{7-t} + 2t >4[/m]
[m]\sqrt{7-t} >4-2t[/m]
[m]\left\{\begin{matrix}4-2t <0 \\7-t ≥0 \end{matrix}\right.[/m] или [m]\left\{\begin{matrix}4-2t ≥ 0 \\(\sqrt{7-t})^2 >(4-2t)^2\end{matrix}\right.[/m]
[m]\left\{\begin{matrix}-2t <-4 \\-t ≥-7 \end{matrix}\right.[/m] или [m]\left\{\begin{matrix}-2t ≥ -4 \\7-t >16-16t+4t^2\end{matrix}\right.[/m]
[m]4t^2-15t+9 <0[/m] D=225-144=81; [m] t_{1}=\frac{15-9}{8}[/m] ;[m] t_{2}=\frac{15+9}{8}[/m]
[m]2 < t ≤ 7[/m] или [m]\frac{3}{4} < t ≤ 2[/m]
14.
ОДЗ: [m]\left\{\begin{matrix}\frac{(x+2)^2\cdot (x+5)}{5}>0 \\\frac{x+5}{5}>0 \end{matrix}\right.[/m]
[m] lg^{2}\frac{(x+2)^2\cdot (x+5)}{5}-lg^{2}\frac{x+5}{5} <0[/m]
[m] (lg\frac{(x+2)^2\cdot (x+5)}{5}-lg\frac{x+5}{5}) \cdot (lg\frac{(x+2)^2\cdot (x+5)}{5}+lg\frac{x+5}{5})<0[/m]
[m]\left\{\begin{matrix}lg\frac{(x+2)^2\cdot (x+5)}{5}-lg\frac{x+5}{5} <0 \\lg\frac{(x+2)^2\cdot (x+5)}{5}+lg\frac{x+5}{5}>0 \end{matrix}\right.[/m] или [m]\left\{\begin{matrix}lg\frac{(x+2)^2\cdot (x+5)}{5}-lg\frac{x+5}{5} >0 \\lg\frac{(x+2)^2\cdot (x+5)}{5}+lg\frac{x+5}{5}<0\end{matrix}\right.[/m]
[m]\left\{\begin{matrix}lg\frac{(x+2)^2\cdot (x+5)}{5}<lg\frac{x+5}{5} \\lg\frac{(x+2)^2\cdot (x+5)}{5}> -lg\frac{x+5}{5} \end{matrix}\right.[/m] или [m]\left\{\begin{matrix}lg\frac{(x+2)^2\cdot (x+5)}{5}>lg\frac{x+5}{5}\\lg\frac{(x+2)^2\cdot (x+5)}{5}<-lg\frac{x+5}{5}\end{matrix}\right.[/m]
[m]\left\{\begin{matrix}lg\frac{(x+2)^2\cdot (x+5)}{5}<lg\frac{x+5}{5} \\lg\frac{(x+2)^2\cdot (x+5)}{5}>lg\frac{5}{x+5} \end{matrix}\right.[/m] или [m]\left\{\begin{matrix}lg\frac{(x+2)^2\cdot (x+5)}{5}>lg\frac{x+5}{5}\\lg\frac{(x+2)^2\cdot (x+5)}{5}<lg\frac{5}{x+5} \end{matrix}\right.[/m]
Логарифмическая функция с основанием 10 > 1 возрастающая, поэтому
[m]\left\{\begin{matrix}\frac{(x+2)^2\cdot (x+5)}{5}<\frac{x+5}{5} \\\frac{(x+2)^2\cdot (x+5)}{5}>\frac{5}{x+5} \end{matrix}\right.[/m] или [m]\left\{\begin{matrix}\frac{(x+2)^2\cdot (x+5)}{5}>\frac{x+5}{5}\\\frac{(x+2)^2\cdot (x+5)}{5}<\frac{5}{x+5} \end{matrix}\right.[/m]
[m]\left\{\begin{matrix}\frac{(x+2)^2\cdot (x+5)}{5}-\frac{x+5}{5} <0 \\\frac{(x+2)^2\cdot (x+5)}{5}-\frac{5}{x+5}>0 \end{matrix}\right.[/m] или [m]\left\{\begin{matrix}\frac{(x+2)^2\cdot (x+5)}{5}-\frac{x+5}{5}>0\\\frac{(x+2)^2\cdot (x+5)}{5}-\frac{5}{x+5} < 0 \end{matrix}\right.[/m]
[m]\left\{\begin{matrix}\frac{x+5}{5}\cdot ((x+2)^2-1)<0 \\\frac{(x+2)^2\cdot (x+5)^2-5^2}{5(x+5)}>0 \end{matrix}\right.[/m] или [m]\left\{\begin{matrix}\frac{x+5}{5}\cdot ((x+2)^2-1)>0\\\frac{(x+2)^2\cdot (x+5)^2-5^2}{5(x+5)}< 0 \end{matrix}\right.[/m]
[m]\left\{\begin{matrix}(x+5)(x+1)(x+3)<0 \\\frac{((x+2)(x+5)-5)((x+2)(x+5)+5)}{5(x+5)}>0 \end{matrix}\right.[/m] или [m]\left\{\begin{matrix}(x+5)(x+1)(x+3)>0\\\frac{((x+2)(x+5)-5)((x+2)(x+5)+5)}{5(x+5)}< 0 \end{matrix}\right.[/m]