Đáp án:
\(\begin{array}{l}
h\left( x \right) > 0 \Leftrightarrow x \in \left( { - \infty ; - 3} \right)\\
h\left( x \right) < 0 \Leftrightarrow x \in \left( { - 3; + \infty } \right)\\
h\left( x \right) = 0 \to x = - 3
\end{array}\)
Giải thích các bước giải:
Xét:
\(\begin{array}{l}
h(x) = {(x + 1)^4} - {(x + 5)^4}\\
h\left( x \right) = 0\\
\to {(x + 1)^4} - {(x + 5)^4} = 0\\
\to {(x + 1)^4} = {(x + 5)^4}\\
\to \left| {x + 1} \right| = \left| {x + 5} \right|\\
\to \left[ \begin{array}{l}
x + 1 = x + 5\\
x + 1 = - x - 5
\end{array} \right.\\
\to \left[ \begin{array}{l}
1 = 5\left( l \right)\\
2x = - 6
\end{array} \right.\\
\to x = - 3
\end{array}\)
BXD:
x -∞ -3 +∞
h(x) + 0 -
\(\begin{array}{l}
KL:h\left( x \right) > 0 \Leftrightarrow x \in \left( { - \infty ; - 3} \right)\\
h\left( x \right) < 0 \Leftrightarrow x \in \left( { - 3; + \infty } \right)\\
h\left( x \right) = 0 \to x = - 3
\end{array}\)
