Đáp án:
$f(x) > 0 \Leftrightarrow \left[\begin{array}{l}-3 < x < -2\\-1 < x < 0\\1 < x < 2\\x > 3\end{array}\right.$
Giải thích các bước giải:
\(\begin{array}{l}
Đặt\ \ f(x) = x^7 -14x^5+49x^3 -36x\\
\ \Leftrightarrow\ f(x) = (x-3)(x-2)(x-1)x(x+1)(x+2)(x+3)\\
\text{Bảng xét dấu:}\\
\begin{array}{|c|ccc|}
\hline
x&-\infty&&-3&&-2&&-1&&0&&1&&2&&3&&+\infty\\
\hline
x+3&&-&\ 0&+&\vert&+&\vert&+&\vert&+&\vert&+&\vert&+&\vert&+&\\
\hline
x+2&&-&\vert&-&\ 0&+&\vert&+&\vert&+&\vert&+&\vert&+&\vert&+&\\
\hline
x+1&&-&\vert&-&\vert&- &\ 0&+&\vert&+&\vert&+&\vert&+&\vert&+&\\
\hline
x&&-&\vert&-&\vert&-&\vert&-& 0&+&\vert&+&\vert&+&\vert&+&\\
\hline
x-1&&-&\vert&-&\vert&-&\vert&-&\vert&-&0&+&\vert&+&\vert&+&\\
\hline
x-2&&-&\vert&-&\vert&-&\vert&-&\vert&-&\vert&-&0&+&\vert&+&\\
\hline
x-3&&-&\vert&-&\vert&-&\vert&-&\vert&-&\vert&-&\vert&-&0&+&\\
\hline
f(x)&&-&0&+&0&-&0&+&0&-&0&+&0&-&0&+&\\
\hline
\end{array}\\
\text{Dựa vào bảng xét dấu, ta được:}\\
\quad f(x) > 0 \Leftrightarrow \left[\begin{array}{l}-3 < x < -2\\-1 < x < 0\\1 < x < 2\\x > 3\end{array}\right.
\end{array}\)
