Question

1/(5-lgx)+2/(1+lgx)=1

Answer 1

$\frac{1}{5-lg(x)} +\frac{2}{1+lg(x)} =1\\\left \{ {{x>0} \atop {x\neq 10^{-1} }}\atop {x\neq 10^{5} }} \right. \\1+lg(x)+2(5-lg(x))-(5-lg(x))(1+lg(x))=0\\1+lg(x)+10-2lg(x)-5-4lg(x)+lg^2(x)=0\\lg^2(x)-5lg(x)+6=0\\(lg(x)-2)(lg(x)-3)=0\\lg(x)=2=>x=10^{2}\\lg(x)=3=>x=10^{3}$

Answer 2

$\frac{1}{5 - log(x) } + \frac{2}{1 + log(x) } = 1 \\ ОДЗ:x > 0 \\ Замена: log(x) = t \\ \frac{1}{5 - t} + \frac{2}{1 + t} = 1 \\ \frac{1 + t + 2(5 - t)}{(5 - t)(1 + t)} - \frac{(5 - t)(1 + t)}{(5 - t)(1 + t)} = 0 \\ \frac{1 + t + 10 - 2t - (5 + 5t - t - {t}^{2} )}{(5 - t)(1 + t)} = 0 \\ \frac{11 - t - 5 - 4t + {t}^{2} }{(5 - t)(1 + t)} = 0 \\ ОДЗ_{t}:t≠5, \: t≠ - 1 \\ {t}^{2} - 5t + 6 = 0 \\ t_{1} = 2 \\ t_{2} = 3 \\ Обратная \: замена: \\ log(x) = 2, \: x = {10}^{2} \\ log(x) = 3, \: x = {10}^{3}$

Ответ: 100, 1000.