区间qūjiān

Khai niem co ban

Mot khoang dai dien cho tap con lien tuc cua so thuc. Ky hieu khoang cung cap phuong phap ngan gon de mo ta pham vi so tren truc so.

Dinh nghia

Khoang la tap hop tat ca cac so thuc giua hai diem cuoi $a$ va $b$, trong do $a \leq b$.

Bon loai co ban

Loai 1: Khoang dong $[a, b]$

$[a, b] = \{x \in \mathbb{R} : a \leq x \leq b\}$

Bao gom ca hai diem cuoi.

Vi du: $[1, 5]$ chua $1, 2, 3, 4, 5$ va tat ca cac so thuc o giua.

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Loai 2: Khoang mo $(a, b)$

$(a, b) = \{x \in \mathbb{R} : a < x < b\}$

Khong bao gom ca hai diem cuoi.

Vi du: $(1, 5)$ chua tat ca cac so giua $1$ va $5$, nhung khong bao gom $1$ va $5$.

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Loai 3: Khoang nua mo $[a, b)$

$[a, b) = \{x \in \mathbb{R} : a \leq x < b\}$

Bao gom diem cuoi trai, khong bao gom diem cuoi phai.

Vi du: $[1, 5)$ bao gom $1$, nhung khong bao gom $5$.

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Loai 4: Khoang nua mo $(a, b]$

$(a, b] = \{x \in \mathbb{R} : a < x \leq b\}$

Khong bao gom diem cuoi trai, bao gom diem cuoi phai.

Vi du: $(1, 5]$ bao gom $5$, nhung khong bao gom $1$.

Khoang vo han

Khoang co the mo rong den vo cuc:

Ky hieu	Mo ta	Ky hieu tap hop
$[a, +\infty)$	Tat ca $x \geq a$	$\{x \in \mathbb{R} : x \geq a\}$
$(a, +\infty)$	Tat ca $x > a$	$\{x \in \mathbb{R} : x > a\}$
$(-\infty, b]$	Tat ca $x \leq b$	$\{x \in \mathbb{R} : x \leq b\}$
$(-\infty, b)$	Tat ca $x < b$	$\{x \in \mathbb{R} : x < b\}$
$(-\infty, +\infty)$	Tat ca so thuc	$\mathbb{R}$

Quan trong: Vo cuc luon duoc viet voi ngoac don vi $\infty$ khong phai la diem cuoi.

Quy tac ngoac

Ngoac	Y nghia	Ky hieu
$[$ hoac $]$	Bao gom diem cuoi	Diem dac
$($ hoac $)$	Khong bao gom diem cuoi	Diem rong

Cac phep toan khoang

Giao

$[1, 5] \cap [3, 7] = [3, 5]$

Cac phan tu chung cua ca hai khoang.

Hop

$[1, 3] \cup [5, 7] = [1, 3] \cup [5, 7]$

Tat ca cac phan tu thuoc it nhat mot khoang.

Phan bu

$[1, 5]^c = (-\infty, 1) \cup (5, +\infty)$

Tat ca cac so thuc khong nam trong khoang.

Bai tap thuc hanh CSCA

1. Viet tap nghiem cua $-2 < x \leq 5$ bang ky hieu khoang.

2. Tinh: $[0, 4] \cap (2, 6]$

3. Xac dinh phan bu cua $(1, 3)$ doi voi $\mathbb{R}$.

4. Neu $A = [-1, 3]$ va $B = (0, 5)$, tim $A \cup B$ va $A \cap B$.

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Dap an:

1. $(-2, 5]$
2. $(2, 4]$
3. $(-\infty, 1] \cup [3, +\infty)$
4. $A \cup B = [-1, 5)$, $A \cap B = (0, 3]$

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Meo hoc tap: Ky hieu khoang la nen tang de hieu mien xac dinh ham so va nghiem bat dang thuc trong cac ky thi CSCA!