Question

Помогите, пожалуйста.

Answer 1

$1)\; \sqrt{log_{x}25+3}= \frac{1}{log_5x} \; ,\; ODZ:\; \left \{ {{x\ \textgreater \ 0,\; log_5x\ne 0} \atop {log_{x}25+3 \geq 0}} \right. \; , \left \{ {{x\ \textgreater \ 0\; ,\; x\ne 1} \atop {\frac{1}{log_{25}x}+3}\geq 0}} \right. \\\\\frac{1}{log_{25}x}+3=\frac{1}{\frac{1}{2}log_5x}+3=\frac{2}{log_5x}+3= \frac{2+3log_5x}{log_5x} \geq 0\; ,\\\\t=log_5x\; ,\; \frac{2+3t}{t} \geq 0\; ,\; \; \; +++(-\frac{2}{3})---(0)+++\; \; \left [ {{t > 0} \atop {t \leq -\frac{2}{3}}} \right.$
 
 $log_5x \leq -\frac{2}{3}\; ,$   $x \leq 5^{-\frac{2}{3}}\; ,\; x \leq \frac{1}{\sqrt[3]{25}}\approx 0,34$

$log_5x > 0\; ,\; x > 1\\\\ODZ:\; \; x\in (0;\frac{1}{\sqrt[3]{25}}\, ]\cup (1,+\infty )\\\\log_{x}25+3=\frac{1}{log^2_5x}\; ,\; \; \frac{2}{log_5x}+3-\frac{1}{log^2_5x}=0\\\\ \frac{2log^2_5x+3log_5x-1}{log^2_5x}=0\; ,\; \; 2log^2_5x+3log_5x-1=0\; \Rightarrow \\\\log_5x=-1\; \; ili\; \; log_5x=\frac{1}{3}\\\\x=5^{-1}=\frac{1}{5}=0,2\; \; ili\; \; x=5^{\frac{1}{3}}=\sqrt[3]5\approx 1,71\\\\Otvet:\; \; 0,2\; \; ili\; \; \sqrt[3]5\; .$

$2)\; \sqrt{2log^2_2x+3log_2x-5}=log_2(2x)\; ,\; \; \; ODZ:\; \; \left \{ {{x\ \textgreater \ 0} \atop {2log^2_2x+3log_2x-5 \geq 0}} \right. \\\\2t^2+3t-5 \geq 0\; ;\; \; \; t_1=-2,5\; ,\; \; t_2=1\\\\+++(-2,5)---(1)+++\; \; \; \; \left [ {{t \geq 1} \atop {t \leq -2,5}} \right. \\\\log_2x \leq -2,5\; ;\; \; x \leq 2^{-2,5}\; ,\; \; x \leq \sqrt{2^{-5}}=\frac{1}{\sqrt{32}}\approx 0,18\\\\log_2x \geq 1\; ,\; \; x \geq 2\\\\ODZ:\; \; x\in (0;\sqrt{2^{-5}}\, ]\cup [2;+\infty )\\\\2log^2_2x+3log_2x-5=(1+log_2x)^2$

$t=log_2x\; ,\; 2t^2+3t-5=1+2t+t^2\\\\t^2+t-6=0\; ,\; \; \; t_1=-3\; ,\; \; t_2=2\\\\log_2x=-3\; ,\; \; x=2^{-3}=\frac{1}{8}=0,125\\\\log_2x=2\; ,\; \; x=2^2=4\\\\Otvet:\; \; 0,125\; \; ili\; \; 4\; .$