Question

Решить неравенство методом интервалов. x^2(x-2)^3(x+3)/(x-4)^7>0

Answer 1

Ответ:

$(-3;0)\cup(0;2)\cup(4;+\infty)$

Объяснение:

$\frac{x^2(x-2)^3(x+3)}{(x-4)^7} > 0$

ОДЗ:

$x\neq 4$

$\displaystyle {\begin{array}{|c|c|c|c|c|c|c|c|c|c|}\cline{1-10} x&(-\infty;-3)&-3&(-3; 0)&0&(0; 2)&2&(2;4)&4&(4;+\infty) \\ \cline{1-10} x^{2}&+&+&+&0&+&+&+&+&+ \\ \cline{1-10} \left(x - 2 \right)^{3}&-&-&-&-&-&0&+&+&+ \\ \cline{1-10} x + 3&-&0&+&+&+&+&+&+&+ \\ \cline{1-10} \left(x - 4 \right)^{7}&-&-&-&-&-&-&-&\times&+ \\ \cline{1-10} \frac{x^2(x-2)^3(x+3)}{(x-4)^7}&-&0&+&0&+&0&-&\times&+ \\ \cline{1-10} \end{array}}$

$\HUGE\frac{}{{\ \ \ \ \ \ \atop}}\!\!\!^{_{^{-3}\atop{\bullet}}}\!\!\!\frac{{_-}}{{\phantom{\ \ \ \ \ \ \atop}}}\!\!^{_{^{0}\atop{{o}}}}\!\!\frac{_-}{{\ \ \ \ \ \ \atop}}\!\!\!^{_{^{2}\atop{\bullet}}}\!\!\!\frac{}{{\ \ \ \ \ \ \atop}}\!\!\!^{_{^{4}\atop{\bullet}}}\!\!\!\frac{{_-}}{{\phantom{\ \ \ \ \ \ \atop}}}\!\!\!^{_{\atop\blacktriangleright}}$

$x\in(-3;0)\cup(0;2)\cup(4;+\infty)$