Question

Помогите пожалуйста !
Любые 3 задачи

Answer 1

$4451.\; \; \; log_{x}13>log_{x}11\; \; ,\; \; \; ODZ:\; x>0\; ,\; x\ne 1\\\\log_{x}13-log_{x}11>0\\\\log_{x}\frac{13}{11}>0\; ,\; \; \frac{1}{log_{\frac{13}{11}}x}>0\; \; \to \; \; log_{\frac{13}{11}}x>0\; \; \to \; \; x>1\\\\Otvet:\; \; x\in (1,+\infty )\, .\\\\4453.\; \; \; log_{x+2}9>log_{x}3\; ,\\\\ODZ:\; \left \{ {{x>0\; ,\; x\ne 1} \atop {x+2>0\; ,\; x+2\ne 1}} \right.\; \left \{ {{x>0\; ,\; x\ne 1} \atop {x>-2\; ,\; x\ne -2}} \right. \; \; \to \; \; x\in (0,1)\cup (1,+\infty )$

$\frac{1}{log_9(x+2)}-\frac{1}{log_3x}>0\; \; ,\; \; \frac{1}{\frac{1}{2}log_3(x+2)}-\frac{1}{log_3x}>0\; \; ,\\\\\frac{2log_3x-log_3(x+2)}{log_3x\cdot log_3(x+2)}>0\; \; \to \; \; \frac{(x^2-x-2)(3-1)}{(3-1)(x-1)(3-1)(x+2-1)}>0\; \; (x>0\; ,\; x\ne 1)\\\\\frac{x^2-x-2}{(x-1)(x+1)}>0\; ,\; \; \frac{(x-2)(x+1)}{(x-1)(x+1)}>0\; ,\; (x\ne -1\; ,\; x>0\; ,\; x\ne 1)\\\\+++(-1)+++(1)---(2)+++\\\\x\in (-\infty ,-1)\cup (-1,1)\cup (2,+\infty )\; \; ,\; (\, x>0\; ,\; x\ne \pm 1\, )\\\\Otvet:\; \; x\in (0,1)\cup (2,+\infty )$

$4459.\; \; \; log_{\frac{2x}{x^2+1} }(5-4x)>-1\\\\ODZ:\; \; \left\{\begin{array}{c}\frac{2x}{x^2+1}>0\\\frac{2x}{x^2+1}\ne 1\\5-4x>0\end{array}\right\; \left\{\begin{array}{c}x>0\\\frac{-(x^2-2x+1)}{x^2+1}\ne 0\\4x<5\end{array}\right\; \left\{\begin{array}{c}x>0\\x\ne 1\\x<\frac{5}{4}\end{array}\right \quad \Rightarrow \\\\x\in (0,1)\cup (1,\frac{5}{4})\\\\log_{\frac{2x}{x^2+1}}(5-4x)+log_{\frac{2x}{x^2+1}}\frac{2x}{x^2+1}}>0\; \; \Leftrightarrow \; \; \left\{\begin{array}{c}x\in (0,1)\cup (1,\frac{5}{4})\\(\frac{2x}{x^2+1}-1)(5-4x-\frac{x^2+1}{2x})>0\end{array}\right$

$\\\\\frac{2x-x^2-1}{x^2+1}\cdot \frac{(5-4x)\cdot 2x-(x^2+1)}{x^2+1}>0\; \; \Rightarrow \; \; \frac{-(x-1)^2}{x^2+1}\cdot \frac{-(9x^2-10x+1)}{2x}>0\\\\9x^2-10x+1>0\; \; pri\; \; x\in R\; \; \; [D<0]\; ,\\\\(x-1)^2>0\; pri\; \; x\ne 1\; \; \; [\, (x-1)^2\geq 0\; ]\; ,\\\\x^2+1>0\; \; pri\; \; x\in R\; \; [\; x^2+1\geq 1\; ]\, .$

$\frac{(x-1)^2}{2x}>0\; \; \Rightarrow \; \; ---(0)+++(1)+++ ,\\\\x\in (0,1)\cup (1,+\infty )\\\\Otvet:\; \; x\in (0,1)\cup (1,\frac{5}{4})\, .$