a) \(A=\left\{x\in R|x-\sqrt[]{3-2x}=0\right\}\)
\(B=\left\{x\in R|x^2+2x-3=0\right\}\)
\(\)\(x-\sqrt[]{3-2x}=0\)
\(\Leftrightarrow\sqrt[]{3-2x}=x\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\3-2x=x^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2+2x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow x=1\)
\(\Rightarrow A=\left\{1\right\}\)
\(x^2+2x-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
\(\Rightarrow B=\left\{-3;1\right\}\)
Vậy \(A\subset B\)
b) \(A=\left\{x\in N|x^2-2x+1>10\right\}\)
\(B=\left\{x\in N|x>=2\right\}\)
\(x^2-2x+1>10\)
\(\Leftrightarrow\left(x-1\right)^2>\left(\sqrt[]{10}\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1< -\sqrt[]{10}\\x-1>\sqrt[]{10}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x< 1-\sqrt[]{10}\\x>1+\sqrt[]{10}\end{matrix}\right.\)
\(\Rightarrow A=(-\infty;1-\sqrt[]{10})\cup(1+\sqrt[]{10};+\infty)\)
\(B=[2;+\infty)\)
mà \(1-\sqrt[]{10}< 2< 1+\sqrt[]{10}\)
Vậy 2 tập hợp không có quan hệ gì giữa nhau
