d) $y= \dfrac{ \left | x+2 \right | + \left | x-2 \right | }{\left | x+1 \right | - \left | x-1 \right | }$
+ TXĐ:
Ta có: $\left | x+1 \right | - \left | x-1 \right | \\
= \left\{\begin{matrix}
-2 \ \ (x \leq -1) \\ (x+1)-(1-x) \ \ (-1 \leq x \leq 1)
\\ 2 \ \ (x \geq 1)
\end{matrix}\right. \\
\\
= \left\{\begin{matrix}
-2 \ \ (x \leq -1) \\ 2x \ \ (-1 \leq x \leq 1)
\\ 2 \ \ (x \geq 1)
\end{matrix}\right.
$
Nên $D=\mathbb{R} \setminus \left\{ 0 \right\}$
$\Rightarrow \forall x \in D, -x \in D$
+ $f(-x)=\dfrac{ \left | -x+2 \right | + \left | -x-2 \right | }{\left | -x+1 \right | - \left | -x-1 \right | } \\
= \dfrac{ \left | 2-x \right | + \left | -(x+2) \right | }{\left | 1-x \right | - \left | -(x+1) \right | } \\
= \dfrac{ \left | -(-2+x) \right | + \left | x+2 \right | }{\left | -(-1+x) \right | - \left | x+1 \right | } \\
= \dfrac{ \left | x-2 \right | + \left | x+2 \right | }{\left | x-1 \right | - \left | x+1 \right | } \\
= - \dfrac{ \left | x+2 \right | + \left | x-2 \right | }{\left | x+1 \right | - \left | x-1 \right | } = -f(x)
$
Vậy hàm $f(x)$ là hàm lẻ
[TẶNG BẠN] TRỌN BỘ Bí kíp học tốt 08 môn
