Xác định các tập hợp sau đây:

* Hướng dẫn giải

a) Ta có: $\left(-\infty ;0\right]=\left\{x\in ℝ|x\le 0\right\}$và $\left[-\pi ;\pi \right]=\left\{x\in ℝ|-\pi \le x\le \pi \right\}$

$\Rightarrow \left(-\infty ;0\right]\cup \left[-\pi ;\pi \right]=\left\{x\in ℝ|x\le 0,-\pi \le x\le \pi \right\}=\left\{x\in ℝ|-\pi \le x\le 0\right\}=\left[-\pi ;0\right].$

$\Rightarrow \left(-\infty ;0\right]\cup \left[-\pi ;\pi \right]$ = {x $\in ℝ$| x ≤ 0 hoặc $-\pi \le x\le \pi$} = $\left\{x\in ℝ|x\le \pi \right\}$= $\left(-\infty ;\pi \right]$.

Vậy $\left(-\infty ;0\right]\cup \left[-\pi ;\pi \right]=\left(-\infty ;\pi \right]$.

b) Ta có: [-3,5; 2] = $\left\{x\in ℝ|-3,5\le x\le 2\right\}$ và (-2; 3,5) = $\left\{x\in ℝ|-2<x<3,5\right\}$

⇒ [-3,5; 2] ∩ (-2; 3,5) = $\left\{x\in ℝ|-2<x\le 2\right\}=\left(-2;2\right]$.

Vậy [-3,5; 2] ∩ (-2; 3,5) = (-2; 2].

c) Ta có $\left(-\infty ;\sqrt{2}\right]=\left\{x\in ℝ|x\le \sqrt{2}\right\}.$ và $\left[1;+\infty \right)=\left\{x\in ℝ|x\ge 1\right\}$

$\Rightarrow \left(-\infty ;\sqrt{2}\right]\cap \left[1;+\infty \right)=\left\{x\in ℝ|1\le x\le \sqrt{2}\right\}=\left[1;\sqrt{2}\right].$

Vậy $\left(-\infty ;\sqrt{2}\right]\cap \left[1;+\infty \right)=\left[1;\sqrt{2}\right].$

d) Ta có $\left(-\infty ;\sqrt{2}\right]=\left\{x\in ℝ|x\le \sqrt{2}\right\}$ và $\left[1;+\infty \right)=\left\{x\in ℝ|x\ge 1\right\}$

$\Rightarrow \left(-\infty ;\sqrt{2}\right]\backslash \left[1;+\infty \right)$ = {$x\in ℝ|x\le \sqrt{2}$ và x < 1} = $\left(-\infty ;1\right)$.

Vậy $\left(-\infty ;\sqrt{2}\right]\backslash \left[1;+\infty \right)=\left(-\infty ;1\right)$.