Question

Решите уравнение пожалуйста!

Answer 1

$frac{1}{lgx + lg0.1} - frac{1}{lgx - lg0.1}= frac{2}{3} \ frac{1}{lgx - 1} - frac{1}{lgx + 1}= frac{2}{3} \ frac{lgx + 1 - (lgx - 1)}{(lgx - 1)(lgx + 1)}= frac{2}{3}\ frac{lgx + 1 - lgx + 1}{(lgx - 1)(lgx + 1)}= frac{2}{3}\ frac{2}{(lg x )^{2} - 1}= frac{2}{3}\ left { {{(lg x )^{2} - 1= frac{2*3}{2}} atop {(lg x)^{2} neq 0}} right. \ left { {{(lg x )^{2} = 4} atop {(lg x)^{2} neq 0}} right. \ lg x = 2; lg x = -2 \ lg x = lg 100; lg x = lg frac{1}{100}$
х=100   или    х=1/100

Ответ:  100;   1/100

Answer 2

$frac{1}{lgx+lg0,1} - frac{1}{lgx-lg0,1}= frac{2}{3}$
$frac{1}{lgx+lg frac{1}{10}} - frac{1}{lgx-lg frac{1}{10} }= frac{2}{3}$
$frac{1}{lgx-1} - frac{1}{lgx+1}= frac{2}{3}$
$frac{lgx+1-lgx+1}{lg^2x-1}= frac{2}{3}$
$frac{2}{lg^2x-1}= frac{2}{3}$
$frac{1}{lg^2x-1}= frac{1}{3}$
Пусть lgx=t
$frac{1}{t^2-1}= frac{1}{3}$
$left[begin{array}{ccc}1*3=t^2-1\t^2-1 neq 0end{array}right$
$left[begin{array}{ccc}t^2=4\t neqб1end{array}right$
$left[begin{array}{ccc}t=б2\t neqб1end{array}right$
lgx=2 и lgx=-2 при х>0
$x_{1}=100$ и $x_{2}= frac{1}{100}$