1111+3333=4444

6666+2222=8888

9999+0000=9999

1111+3333= 4444

6666+2222=8888

9999+0000=9999

chúc b học tốt

1111 + 3333 = 4444

6666 + 2222 = 8888

9999 + 0000 = 9999

\(A=\dfrac{2}{3}+\dfrac{2}{3^2}+\dfrac{2}{3^3}+....+\dfrac{2}{3^{2023}}\)

\(3A=2+\dfrac{2}{3}+\dfrac{2}{3^2}+....+\dfrac{2}{3^{2022}}\)

\(3A-A=\left(2+\dfrac{2}{3}+\dfrac{2}{3^2}+...+\dfrac{2}{3^{2022}}\right)-\left(\dfrac{2}{3}+\dfrac{2}{3^2}+....+\dfrac{2}{3^{2023}}\right)\)

\(2A=2-\dfrac{2}{3^{2023}}\)

\(A=\left(2-\dfrac{2}{3^{2023}}\right)\times\dfrac{1}{2}\)

\(A=2\times\dfrac{1}{2}-\dfrac{2}{3^{2023}}\times\dfrac{1}{2}\)

\(A=1-\dfrac{1}{3^{2023}}\)

=> \(A< 1\left(đpcm\right)\)

`2/(1xx3)+2/(3xx5)+2/(5xx7)+...+2/(99xx101)` đề phải ntn chứ mà nhỉ

`=1/1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101`

`=1/1-1/101`

`=101/101-1/101`

`=100/101`

(Sửa phần 3 / 3 x 5 = 2 / 3 x 5)

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{99\times101}\)

Ta có: \(=2\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{99\times101}\right)\)

\(=2\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=2\times\left(1-\dfrac{1}{101}\right)\)

\(=2\times\dfrac{100}{101}\)

\(=\dfrac{200}{101}\)

Sửa bài ( dòng 3 đến hết bài )

... = \(2\times\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)

\(=1-\dfrac{1}{101}\)

\(=\dfrac{100}{101}\)

N=1/2+1/2²+...+1/2¹⁰

2N=1+1/2+...+1/2⁹

2N-N=1-1/2¹⁰=1-1/1024=1023/1024

Giải:

N=1/2+1/2²+1/2³+...+1/2⁹+1/2¹⁰

2N=1+1/2+1/2²+...+1/2⁸+1/2⁹

2N-N=(1+1/2+1/2²+...+1/2⁸+1/2⁹)-(1/2+1/2²+1/2³+...+1/2⁹+1/2¹⁰)

N=1-1/2¹⁰=1023/1024

Chúc bạn học tốt!

A=\(\dfrac{3}{1}\).(\(\dfrac{3}{2.5}\)+\(\dfrac{3}{5.8}\)+...+\(\dfrac{3}{98.101}\))

A=3.(\(\dfrac{1}{2}\)-\(\dfrac{1}{5}\)+\(\dfrac{1}{5}\)-\(\dfrac{1}{8}\)+...+\(\dfrac{1}{98}\)-\(\dfrac{1}{101}\))

A=3.(\(\dfrac{1}{2}\)-\(\dfrac{1}{101}\))

A=3.\(\dfrac{98}{202}\)

A=\(\dfrac{294}{202}\)=\(\dfrac{147}{101}\)

Đặt \(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{99^2}+\dfrac{1}{100^2}\)

\(< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)

\(B=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{98.99}+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{98}-\dfrac{1}{99}+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\) \(\Rightarrow A< \dfrac{99}{100}\)

\(1-\dfrac{1}{2^2}-\dfrac{1}{3^2}-...-\dfrac{1}{100^2}=1-\left(\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{100^2}\right)=1-A>\dfrac{1}{100}\)

\(x=\dfrac{29}{24}\)

X : 1/2 = 29/12

X         = 29/12 x 2/1 

X         = 29/6

\(x:\dfrac{1}{2}=\dfrac{3}{4}+\dfrac{5}{3}\)

\(x:\dfrac{1}{2}=\dfrac{29}{12}\)

      \(x=\dfrac{29}{12}\)x\(\dfrac{1}{2}\)

      \(x=\dfrac{29}{24}\)

A = \(\dfrac{\dfrac{2022}{1}+\dfrac{2021}{2}+\dfrac{2020}{3}+...+\dfrac{1}{2022}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}}\)

Xét TS = \(\dfrac{2022}{1}\) + \(\dfrac{2021}{2}\) \(\dfrac{2020}{3}\) +... + \(\dfrac{1}{2022}\)

      TS = (1 + \(\dfrac{2021}{2}\)) + (1 + \(\dfrac{2020}{3}\)) + ... + ( 1 + \(\dfrac{1}{2022}\)) + 1 

      TS = \(\dfrac{2023}{2}\) + \(\dfrac{2023}{3}\) +...+ \(\dfrac{2023}{2022}\) + \(\dfrac{2023}{2023}\)

      TS =  2023.(\(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{4}\) +...+ \(\dfrac{1}{2023}\))

A = \(\dfrac{2023.\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\right)}{\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{2023}\right)}\)

 A = 2023

Em cảm ơn ạ