Question

lg(x^2-2)=lg x. Решите пожалуйста!!

Answer 1

$lg(x^2-2)=lgx\\ \left \{ {{x^2-2\ \textgreater \ 0} \atop {x\ \textgreater \ 0}} \right. =\ \textgreater \ \left \{ {{(x- \sqrt{2})(x+ \sqrt{2})\ \textgreater \ 0 } \atop {x\ \textgreater \ 0}} \right. =\ \textgreater \ \left \{ {{x\ \textless \ - \sqrt{2},\; \; x\ \textgreater \ \sqrt{2} } \atop {x\ \textgreater \ 0}} \right. =\ \textgreater \ x\ \textgreater \ \sqrt{2}\\\\x^2-2=x\\x^2-x-2=0\\(x+1)(x-2)=0\\x_1=-1\; \; (\ \textless \ 0)\\x_2=2\; \; (\ \textgreater \ \sqrt{2}) \\\\x=2$

Answer 2

lg (x² - 2) = lgx
ОДЗ: 1) х > 0;
2) x² - 2 > 0 → x² > 2 → x< - √2 ≈ - 1,41; x > √2 ≈ 1,41
Итого ОДЗ х > 1.41
Решаем уравнение
х² - 2 = х
х² - х - 2 = 0
D = 1 + 8 = 9
х1 = 0,5(1 - 3) = -1 не подходит из-за ОДЗ
х2 = 0,5(1 + 3) = 2
Ответ: х = 2