\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{x\left(x+1\right)}=\dfrac{2022}{2023}\)
\(\Rightarrow1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}=\dfrac{2022}{2023}\)
\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{2022}{2023}\)
\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{2022}{2023}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{2023}\)
\(\Rightarrow x+1=2023\)
\(\Rightarrow x=2022\)
Vậy x = 2022
#kễnh

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{x.\left(x+1\right)}\)

= \(\dfrac{2-1}{1.2}+\dfrac{3-2}{2.3}+...+\dfrac{x+1-x}{x.\left(x+1\right)}\)

= \(\dfrac{2}{1.2}-\dfrac{1}{1.2}+\dfrac{3}{2.3}-\dfrac{2}{2.3}+...+\dfrac{x+1}{x.\left(x+1\right)}-\dfrac{x}{x.\left(x+1\right)}\)

= \(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\)

= \(1-\dfrac{1}{x+1}\) =\(\dfrac{2022}{2023}\)

= \(\dfrac{2023}{2023}-\dfrac{1}{x+1}=\dfrac{2022}{2023}\)

⇒ \(x+1=2023\)

\(x=2023-1=2022\)

Bạn cho mình biết là 4 hình vuông có bằng nhau không ạ

Chu vi 1 hình vuông:
\(144:4=36\left(cm\right)\)
Cạnh của 1 hình vuông:
\(36:4=9\left(cm\right)\)
Diện tích 1 hình vuông:
\(9\times9=81\left(cm^2\right)\)
Diện tích của 4 hình vuông:
\(81\times4=324\left(cm^2\right)\)
Đáp số: 324 cm²
#kễnh

27 . 121 - 87 . 27 + 73 .54

= 27 . 121 - 87 . 27 + 73 . 2 . 27

= 27 . (121 - 87 + 73 . 2)

= 27 . 180

= 4860

\(27\cdot121-87\cdot27+73\cdot54\)

\(=27\cdot121-87\cdot27+146\cdot27\)

\(=27\left(121-87+146\right)\)

\(=27\cdot180\)

\(=4860\)

\(x\cdot2^2=2^7\)

\(\Rightarrow x=\dfrac{2^7}{2^2}\)

\(\Rightarrow x=2^{7-2}\)

\(\Rightarrow x=2^5\)

\(\Rightarrow x=32\)

\(\left(x-1\right).3=15\\ =>x-1=5\\ =>x=6\)

\(20:\left(x+1\right)-3=2\\ =>20:\left(x+1\right)=5\\ =>x+1=4\\ =>x=3\)

\(\left(x-1\right).3=15\)

\(x-1=15:3\)

\(x-1=5\)

\(x=5+1\)

\(x=6\)

Vậy \(x=6\)

\(20:\left(x+1\right)-3=2\)

\(\left(x+1\right)-3=20:2\)

\(\left(x+1\right)-3=10\)

\(x+1=10-3\)

\(x+1=7\)

\(x=7-1\)

\(x=6\)

Vậy \(x=6\)

\(\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{402}+5^{403}+5^{404}\right)\\ =31+5^3.\left(1+5+5^2\right)+...+5^{402}.\left(1+5+5^2\right)\\ =31+5^3.31+...+5^{402}.31\\ =31.\left(1+5^3+...+5^{402}\right)⋮31\left(DPCM\right)\)

dpcp là gì vậy ạ

\(x^3-20=7\\ =>x^3=27=3^3\\ =>x=3\)

`x^3-20=7`

`=>x^3=7+20`

`=>x^3=27`

`=>x^3=3^3`

`=>x=3`

Giúp minh với nha 😉

Vì \(m,n\in A\) \(\Rightarrow B\in A\)

Vậy \(m=7\) hoặc \(m=5\)

\(n=5\) hoặc \(n=7\)

\(33\times77+66\times77+77\)

\(=77\times\left(33+66+1\right)\)

\(=77\times100\)

\(=7700\)

1+3+5+7+...+2013
số số hang : (2013-1):2+1=1007
Vậy Ta có tổng là : (2013+1).1007:2=1014049

Số số hạng:

(2013 - 1) : 2 + 1 = 1007 (số)

1 + 3 + 5 + 7 + ... + 2013

= (2013 + 1) . 1007 : 2

= 1014049