Nhiệt độ ở Ottawa lúc 10 giờ là:

\(-4+6=2\left(^0C\right)\)

Đáp số: \(2^0C\)

2 độ C chăng? Vì -4 độ C + 6 độ C = 2 độ C

\(Ư\left(12\right)=\left\{1;2;3;4;6;12;-1;-2;-3;-4;-6;-12\right\}\)

Mà \(2< x< 8\Rightarrow x=\left\{3;4;6\right\}\)

Vậy \(x=\left\{3;4;6\right\}\)

 x = {3;4;6}

Bài 1: 

a. $1991\equiv 1\pmod {10}$
$\Rightarrow 1991^{1997}\equiv 1^{1997}\equiv 1\pmod {10}$
$1997^2\equiv -1\pmod {10}$

$\Rightarrow 1997^{1996}=(1997^2)^{998}\equiv (-1)^{998}\equiv 1\pmod {10}$

Do đó:

$1991^{1997}-1997^{1996}\equiv 1-1\equiv 0\pmod {10}$
Vậy $1991^{1997}-1997^{1996}\vdots 10$

b.

$2^9+2^{99}=2^9(1+2^{90})$

Hiển nhiên $2^9(1+2^{90}\vdots 4$ nên để cm nó chia hết cho 100 thì ta chỉ cần cm $2^{90}+1\vdots 25$

Có:

$2^{10}\equiv -1\pmod {25}$

$\Rightarrow 2^{90}=(2^{10})^9\equiv (-1)^9\equiv -1\pmod {25}$

$\Rightarrow 2^{90}+1\equiv -1+1\equiv 0\pmod {25}$

Vậy $2^{90}+1\vdots 25$

$\Rightarrow 2^9+2^{99}=2^9(2^{90}+1)\vdots 100$

Bài 2:

$x\not\vdots 3$

Tức là $x\equiv \pm 1\pmod 3$

$\Rightarrow x^2\equiv (\pm 1)^2\equiv 1\pmod 3$

\(A=3.2^2+4.2^3+...+60.2^{59}+61.2^{60}\)

\(\Rightarrow2A=3.2^3+4.2^4+...+60.2^{60}+61.2^{61}\)

\(\Rightarrow A-2A=3.2^2+2^3+2^4+...++2^{60}-61.2^{61}\)

\(\Rightarrow-A=5+1+2^1+2^2+2^3+...+2^{60}-61.2^{61}\)

\(\Rightarrow-2A=10+2^1+2^2+2^3+...+2^{61}-61.2^{62}\)

\(\Rightarrow-A-\left(-2A\right)=-4-62.2^{61}+61.2^{62}\)

\(\Rightarrow A=-4+2^{61}\left(-62+61.2\right)\)

\(\Rightarrow A=60.2^{61}-4\)

Lời giải:

$A=2+3.2^2+4.2^3+5.2^4+....+61.2^{60}$

$2A=4+3.2^3+4.2^4+5.2^5+....+61.2^{61}$

$\Rightarrow A=2A-A=2-12-(2^3+2^4+2^5+....+2^{60})+61.2^{61}$

$\Rightarrow A=61.2^{61}-(2^3+2^4+2^5+...+2^{60})-10$

$2A=61.2^{62}-(2^4+2^5+2^6+...+2^{61})-20$

$2A-A=[61.2^{62}-(2^4+2^5+2^6+...+2^{61})-20]-[61.2^{61}-(2^3+2^4+2^5+...+2^{60})-10]$

$\Rightarrow A=61.2^{62}-61.2^{61}-2^{61}-20+2^3+10$

$\Rightarrow A=61.2^{62}-62.2^{61}-2$

$\Rightarrow A=2^{61}(61.2-62)-2=60.2^{61}-2$

a; P = \(\dfrac{-9}{n+1}\) (đk n ≠ -1)

  P  \(\in\) Z ⇔ - 9 ⋮ n + 1

  ⇒ n + 1 \(\in\) Ư(-9) = {-9; -3; -1; 1; 3; 9}

 Lập bảng ta có:

Theo bảng trên ta có:

n \(\in\) {-10; -4; -2; 0; 2; 8}

Kết luận: P = \(\dfrac{-9}{n+1}\) nguyên khi n \(\in\) {-10; -4; -2; 0; 2; 8}

CCô ơi giúp con câu b, c nữa ạ

A = \(\dfrac{2}{2.3}\) + \(\dfrac{2}{3.4}\) + \(\dfrac{2}{4.5}\) + ... + \(\dfrac{2}{199.200}\)

A = 2. (\(\dfrac{1}{2.3}\) + \(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + ... + \(\dfrac{1}{199.200}\))

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) + ... + \(\dfrac{1}{199}\) - \(\dfrac{1}{200}\))

A = 2.(\(\dfrac{1}{2}\) - \(\dfrac{1}{200}\))

A = 2. \(\dfrac{99}{200}\)

A = \(\dfrac{99}{100}\)

Lời giải:
Gọi tổng trên là $A$

$A=2(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{199.200})$

$=2(\frac{3-2}{2.3}+\frac{4-3}{3.4}+\frac{5-4}{4.5}+...+\frac{200-199}{199.200})$

$=2(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....+\frac{1}{199}-\frac{1}{200})$

$=2(\frac{1}{2}-\frac{1}{200})=1-\frac{1}{100}=\frac{99}{100}$

17.230-58.62+17.70+58.362

= 17. (230+70) + 58. (362 - 62)

= 17. 300 + 58 . 300

= (17+58) . 300

= 75 . 300

= 22 500

Đề bài hỏi chứng minh hay làm gì hả bạn?

S=5+5²+5³+...+5²⁰²⁰+5²⁰²¹

5S=5²+5³+5⁴+...+5²⁰²²

5S-S=(5+5²+5³+...+5²⁰²⁰+5²⁰²¹)-(5²+5³+5⁴+...+5²⁰²²)

4S=5-5²⁰²²

S=(5-5²⁰²²):4