Violympic toán 6

( 19x + 2.52 ) : 14 = ( 13 - 8 )2- 4 2

( 19x + 2.52 ) : 14 = 52- 4 2

( 19x + 2.52 ) : 14 = 25- 16

( 19x + 2.52 ) : 14 =9

( 19x + 50 ) : 14 =9

 19x + 50=9 . 14

 19x + 50=126

19x=126-50

19x=76

x=76:19

x=4

(19x + 2.5²) : 14 = (13 - 8)² - 4²

(19x + 2.25) : 14 = 5² - 16

(19x + 50) : 14 = 25 - 16

(19x + 50) : 14 = 9

19x + 50 = 9 . 14

19x + 50 = 126

19x = 126 - 50

19x = 76

x = 76 : 19

x = 4

a: 1-2+3-4+...+99-100

=(1-2)+(3-4)+...+(99-100)

=(-1)+(-1)+...+(-1)

=-1*50=-50

c: 1+2-3-4+....+97+98-99-100

=(1+2-3-4)+(5+6-7-8)+...+(97+98-99-100)

=(-4)+(-4)+...+(-4)

=(-4)*25=-100

\(2023-\left(374+2023\right)+\left(-2024+374\right)\)

\(=2023-374-2023+\left(-2024\right)+374\)

\(=\left(2023-2023\right)+\left(-374+374\right)+\left(-2024\right)\)

\(=0+0+\left(-2024\right)\)

\(=-2024\)

2023-(374+2023)+(-2024+374)

=2023-374-2023-2024+374

=(2023-2023)-(-374+374)-2024

=0-0-2024

=-2024

`#3107.101107`

\(2023-(374 + 2023) + (-2024+374)\)

`= 2023 - 374 - 2023 - 2024 + 374`

`= (2023 - 2023) + (-374 + 374) - 2024`

`= -2024`

TH1: p=3

p+2=3+2=5; p+4=3+4=7

=>Nhận

TH2: p=3k+1

\(p+2=3k+1+2=3k+3=3\left(k+1\right)⋮3\)

=>loại

TH3: p=3k+2

\(p+4=3k+2+4=3k+6=3\left(k+2\right)⋮3\)
=>Loại

Vậy: p=3

số 3 vì 3+2=5,3+4=7

5 và 3 là số nguyên tố

5². 5³ không chia hết cho 3 nhé bạn! Đề sai thì phải

\(\left(x^3-1\right)^{10}+5\left(x^2-x\right)^{20}-7\)

Ta có:

\(\left(x^3-1\right)^{10}\ge0\) với mọi \(x\in R\)

\(\left(x^2-x\right)^{20}\ge0\) với mọi \(x\in R\)

\(\Rightarrow5\left(x^2-x\right)^{20}\ge0\) với mọi \(x\in R\)

\(\Rightarrow\left(x^3-1\right)^{10}+5\left(x^2-x\right)^{20}\ge0\) với mọi \(x\in R\)

\(\Rightarrow\left(x^3-1\right)^{10}+5\left(x^2-x\right)^{20}-7\ge-7\) với mọi \(x\in R\)

Vậy GTNN của \(A\) là \(-7\) khi \(x=1\)

ngầu thí để tên tiếng anh

gõ ko dấu ai dịch được

hok dich dc:))))))

\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)...\left(1-\dfrac{1}{2010}\right)\)

\(=\left(\dfrac{2-1}{2}\right)\left(\dfrac{3-1}{3}\right)\left(\dfrac{4-1}{4}\right)...\left(\dfrac{2010-1}{2010}\right)\)

\(=\dfrac{1.2.3...2009}{2.3.4...2010}\)

\(=\dfrac{1}{2010}\)

\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\dots\left(1-\dfrac{1}{2009}\right)\cdot\left(1-\dfrac{1}{2010}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\cdot\cdot\dfrac{2008}{2009}\cdot\dfrac{2009}{2010}\)

\(=\dfrac{1\cdot2\cdot3\dots2008\cdot2009}{2\cdot3\cdot4\dots2009\cdot2010}\)

\(=\dfrac{1}{2010}\)

1b

2c

3b

4a

5d

6d

7c

8a

9c

10b

nhớ tick cho mik nha.thanks