Đáp án:
\(\begin{array}{l}
1)\quad \text{$y$ là hàm chẵn}\\
2)\quad \text{$y$ là hàm lẻ}\\
3)\quad \text{$y$ là hàm chẵn}\\
4)\quad \text{$y$ là hàm chẵn}\\
5)\quad \text{$y$ là hàm chẵn}\\
6)\quad \text{$y$ là hàm chẵn}\\
7)\quad \text{$y$ là hàm lẻ}\\
8)\quad \text{$y$ là hàm chẵn}\\
9)\quad \text{$y$ là hàm lẻ}\\
10)\quad \text{$y$ là hàm chẵn}\\
11)\quad \text{$y$ là hàm chẵn}\\
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\quad y = f(x) = 3x^2 - 1\\
TXD: D = \Bbb R\\
\forall x\in D \longrightarrow - x\in D\\
\text{Ta có:}\\
f(-x) = 3.(-x)^2 - 1\\
\kern30pt = 3x^2 - 1\\
\kern30pt= f(x)\\
\text{Vậy $y$ là hàm chẵn}\\
2)\quad y = f(x) = \dfrac{x}{x^2 + 1}\\
TXD: D = \Bbb R\\
\forall x\in D \longrightarrow - x\in D\\
\text{Ta có:}\\
f(-x) = \dfrac{-x}{(-x)^2 + 1}\\
\kern30pt = - \dfrac{x}{x^2 + 1}\\
\kern30pt=- f(x)\\
\text{Vậy $y$ là hàm lẻ}\\
3)\quad y = f(x) =-4x^2 + 5|x| - 3\\
TXD: D = \Bbb R\\
\forall x\in D \longrightarrow - x\in D\\
\text{Ta có:}\\
f(-x) = -4.(-x)^2 + 5|-x| - 3\\
\kern30pt =-4x^2 + 5|x| - 3\\
\kern30pt= f(x)\\
\text{Vậy $y$ là hàm chẵn}\\
4)\quad y = f(x) = \dfrac{x^2 + 4}{x^4}\\
TXD: D = \Bbb R\backslash\{0\}\\
\forall x\in D \longrightarrow - x\in D\\
\text{Ta có:}\\
f(-x) =\dfrac{(-x)^2 + 4}{(-x)^4}\\
\kern30pt = \dfrac{x^2 + 4}{x^4}\\
\kern30pt= f(x)\\
\text{Vậy $y$ là hàm chẵn}\\
5)\quad y = f(x) =\dfrac{-x^4 + x^2 + 1}{x}\\
TXD: D = \Bbb R\backslash\{0\}\\
\forall x\in D \longrightarrow - x\in D\\
\text{Ta có:}\\
f(-x) = \dfrac{-(-x)^4 + (-x)^2 + 1}{-x}\\
\kern30pt = -\dfrac{-x^4 + x^2 + 1}{x}\\
\kern30pt= f(x)\\
\text{Vậy $y$ là hàm chẵn}\\
6)\quad y = f(x) = \dfrac{3x^2}{2-|x|}\\
TXD: D = \Bbb R\backslash\{\pm 2\}\\
\forall x\in D \longrightarrow - x\in D\\
\text{Ta có:}\\
f(-x) = \dfrac{3.(-x)^2}{2 - |-x|}\\
\kern30pt = \dfrac{3x^2}{2 - |x|}\\
\kern30pt= f(x)\\
\text{Vậy $y$ là hàm chẵn}\\
7)\quad y = f(x) = \sqrt{x+3} - \sqrt{3 - x}\\
TXD: D = [-3;3]\\
\forall x\in D \longrightarrow - x\in D\\
\text{Ta có:}\\
f(-x) = \sqrt{-x + 3} - \sqrt{3 - (-x)}\\
\kern30pt = \sqrt{3 - x} - \sqrt{x + 3}\\
\kern30pt = - \left(\sqrt{x+3} - \sqrt{3 - x}\right)\\
\kern30pt= -f(x)\\
\text{Vậy $y$ là hàm lẻ}\\
8)\quad y = f(x) = |3x - 2| + |3x + 2|\\
TXD: D = \Bbb R\\
\forall x\in D \longrightarrow - x\in D\\
\text{Ta có:}\\
f(-x) = |3.(-x) - 2| + |3.(-x) + 2|\\
\kern30pt = |3x + 2| + |3x - 2|\\
\kern30pt= f(x)\\
\text{Vậy $y$ là hàm chẵn}\\
9)\quad y = f(x) =|x-2| - |x+2|\\
TXD: D = \Bbb R\\
\forall x\in D \longrightarrow - x\in D\\
\text{Ta có:}\\
f(-x) = |-x - 2| - |-x + 2|\\
\kern30pt = |x + 2| - |x - 2|\\
\kern30pt = - \left(|x -2| - |x + 2|\right)\\
\kern30pt= -f(x)\\
\text{Vậy $y$ là hàm lẻ}\\
10)\quad y = f(x) = \dfrac{|x+1| +|x-1|}{2013x}\\
TXD: D = \Bbb R\backslash\{0\}\\
\forall x\in D \longrightarrow - x\in D\\
\text{Ta có:}\\
f(-x) =\dfrac{|-x + 1|+|-x - 1|}{2013.(-x)} \\
\kern30pt = - \dfrac{|x-1| + |x + 1|}{2013x}\\
\kern30pt= f(x)\\
\text{Vậy $y$ là hàm chẵn}\\
11)\quad y = f(x) = \dfrac{3x^4 - x^2 + 5}{|x|-1}\\
TXD: D = \Bbb R\backslash\{\pm 1\}\\
\forall x\in D \longrightarrow - x\in D\\
\text{Ta có:}\\
f(-x) = \dfrac{3.(-x)^4 - (-x)^2 + 5}{|-x| - 1}\\
\kern30pt = \dfrac{3x^4 - x^2 + 4}{|x| - 1}\\
\kern30pt= f(x)\\
\text{Vậy $y$ là hàm chẵn}\\
\end{array}\)
