Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
a,\\
x\left( {x + 1} \right)\left( {x - \sqrt 2 } \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x + 1 = 0\\
x - \sqrt 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = - 1\\
x = \sqrt 2
\end{array} \right.\\
x \in R \Rightarrow A = \left\{ {0; - 1;\sqrt 2 } \right\}\\
b,\\
\left( {{x^2} + x - 2} \right)\left( {2{x^2} - 8} \right) = 0\\
\Leftrightarrow \left( {x + 2} \right)\left( {x - 1} \right).2.\left( {{x^2} - 4} \right) = 0\\
\Leftrightarrow \left( {x + 2} \right)\left( {x - 1} \right).\left( {x - 2} \right)\left( {x + 2} \right) = 0\\
\Leftrightarrow \left( {x - 1} \right)\left( {x - 2} \right){\left( {x + 2} \right)^2} = 0\\
\Leftrightarrow \left[ \begin{array}{l}
x - 1 = 0\\
x - 2 = 0\\
x + 2 = 0
\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}
x = 1\\
x = 2\\
x = - 2
\end{array} \right.\\
x \in R \Rightarrow B = \left\{ {1;2; - 2} \right\}\\
d,\\
n \in N,\,\,n < \frac{1}{9} \Rightarrow n = 0 \Rightarrow x = \frac{1}{2}n = 0\\
\Rightarrow D = \left\{ 0 \right\}\\
e,\\
\sqrt {{x^2} - 2x - 3} = 1 - x\\
DKXD:\,\,\,{x^2} - 2x - 3 \ge 0 \Leftrightarrow \left( {x - 3} \right)\left( {x + 1} \right) \ge 0 \Leftrightarrow \left[ \begin{array}{l}
x \ge 3\\
x \le - 1
\end{array} \right.\\
\sqrt {{x^2} - 2x - 3} = 1 - x\\
\Leftrightarrow \left\{ \begin{array}{l}
1 - x \ge 0\\
{x^2} - 2x - 3 = {\left( {1 - x} \right)^2}
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le 1\\
{x^2} - 2x - 3 = {x^2} - 2x + 1
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x \le 1\\
- 3 = 1
\end{array} \right.\,\,\,\,\,\left( {vn} \right)\\
\Rightarrow E = \emptyset
\end{array}\)
Em xem lại đề câu C nhé
