Question

Помогите, пожалуйста.

Answer 1

$1); sqrt{log_{x}25+3}= frac{1}{log_5x} ; ,; ODZ:; left { {{x textgreater 0,; log_5xne 0} atop {log_{x}25+3 geq 0}} right. ; , left { {{x textgreater 0; ,; xne 1} atop {frac{1}{log_{25}x}+3}geq 0}} right. \\frac{1}{log_{25}x}+3=frac{1}{frac{1}{2}log_5x}+3=frac{2}{log_5x}+3= frac{2+3log_5x}{log_5x} geq 0; ,\\t=log_5x; ,; frac{2+3t}{t} geq 0; ,; ; ; +++(-frac{2}{3})---(0)+++; ; left [ {{t > 0} atop {t leq -frac{2}{3}}} right.$
 
 $log_5x leq -frac{2}{3}; ,$   $x leq 5^{-frac{2}{3}}; ,; x leq frac{1}{sqrt[3]{25}}approx 0,34$

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

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

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