Найдите наименьшее натуральное значение x,удовлетворяющее системе неравенств [latex] \left \{ {{lg ^{2} x+lg 0,01x\ \textgreater \ 0} \atop { \frac{1}{x}\ \textless \ 1000 }} \right. [/latex] 1) 1; 2) 2; 3) 7; 4) 10; 5)11;

[latex] \left \{ {{lg^2x+lg0.01x\ \textgreater \ 0} \atop { \frac{1}{x}\ \textless \ 1000 }} \right. [/latex] Решаем первое неравенство  [latex]lg^2x+lg0.01x\ \textgreater \ 0 \\ lg^2x + lgx + lg0.01\ \textgreater \ 0 \\ lgx^2 + lgx + lg10^-2 \ \textgreater \ 0 \\ lgx^2+lgx-2\ \textgreater \ 0[/latex] [latex]lgx = t \\ t^2 + t - 2 \ \textgreater \ 0 \\ t^2 + t - 2 = 0 \\ D = 1 + 4*2 = 1 + 8 = 9 \\ t_{1} = \frac{-1+3}{2} = \frac{2}{2} [/latex] [latex] t_{2} = \frac{-1-3}{2} = - \frac{4}{2} = -2 [/latex] ///////-2____1/////// t∈ (-∞; -2) ∨ (1; +∞) или t < -2; t > 1 lg x < -2;     lgx > 1 lgx 10   x < 0.01               x > 10 ОДЗ х > 0 ///////0//0.01_____10/////////         \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ x ∈ (0; 0.01) ∨ (10; +∞) Решаем второе неравенство: [latex] \frac{1}{x} \ \textless \ 1000 \\ \frac{1}{x} - 1000 \ \textless \ 0 \\ \frac{1-1000x}{x}\ \textless \ 0 \\ f(x) = x(1 - 1000x)\ \textless \ 0 [/latex] _-__0___+___0.001__-__ x ∈ (-∞; 0) ∨ (0.001; +∞)  Система: {x ∈ (0; 0.01) ∨ (10; +∞) {x ∈ (-∞; 0) ∨ (0.001; +∞)  Решение: х ∈ (0.001; 0.01) ∨ (10; +∞) Ответ: 5) 11