Đáp án:
\(\begin{array}{l}
h)\quad D = (-2;1)\cup (1;2)\cup (2;+\infty)\\
i)\quad D = [1;+\infty)\backslash\{4\}\\
j)\quad D = \Bbb R\\
k)\quad D = \Bbb R\backslash\left\{\dfrac12\right\}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
h)\quad y = \dfrac{1}{(x^2 -3x + 2)\sqrt{x+2}}\\
DKXD:\begin{cases}x^2 - 3x +2 \ne 0\\x + 2 >0\end{cases}\Leftrightarrow \begin{cases}x \ne 1\\x \ne 2\\x > -2\end{cases}\\
\Rightarrow D = (-2;1)\cup (1;2)\cup (2;+\infty)\\
i)\quad y = \dfrac{1}{\sqrt{x+5} - \sqrt{3x-3}}\\
DKXD: \begin{cases}x + 5 \geqslant 0\\3x - 3 \geqslant 0\\\sqrt{x+5} \ne \sqrt{3x - 3}\end{cases}\Leftrightarrow \begin{cases}x \geqslant -5\\x \geqslant 1\\x \ne 4\end{cases}\\
\Rightarrow D = [1;+\infty)\backslash\{4\}\\
j)\quad y = \dfrac{2x+1}{x^2 + x + 1}\\
DKXD: x^2 + x + 1 \ne 0 \Leftrightarrow \left(x + \dfrac12\right)^2 + \dfrac34 \ne 0\quad \text{(luôn đúng)}\\
\Rightarrow D = \Bbb R\\
k)\quad y = \dfrac{2x-1}{4x^2 - 4x + 1}\\
DKXD: 4x^2 - 4x + 1 \ne 0 \Leftrightarrow (2x - 1)^2 \ne 0 \Leftrightarrow x \ne \dfrac12\\
\Rightarrow D = \Bbb R\backslash\left\{\dfrac12\right\}
\end{array}\)
