cái nì mik chịu

M=(1/5+1/5^2+1/5^3+...+1/5^2023) + 1/5x(1/5+1/5^2+1/5^3+...+1/5^2022) + ... + 1/5^2021x(1/5+1/5^2) + 1/5^2022x1/5

Xét biểu thức N=1/5+1/5^2+1/5^3 + ... + 1/5^k (K>0, k thuộc Z)

=> 5N=1+1/5+1/5^2+1/5^3+...+1/5^(k-1)

=> 4N= 5N - N =1 - 1/5^k

=> 1/5+1/5^2+1/5^3 + ... + 1/5^k = 1/4x(1-1/5^k)

Thay vào biểu thức M, ta có:

M= 1/4x(1-1/5^2023) + 1/5x1/4x(1-1/5^2022) + ... + 1/5^2021x1/4x(1-1/5^2) + 1/5^2022x1/4x(1-1/5)

=> 4M = (1+1/5+1/5^2+...+1/5^2022) - 2023/5^2023

=> 4M = 5/4x(1-1/5^2023)-2023/5^2023 < 5/4

=> M < 5/16 < 1/3 

Vậy M < 1/3 [ vượt chỉ tiêu nhé =)) ]

a:

Sửa đề: \(S=1-3+5-7+...+2021-2023+2025\)

Từ 1 đến 2025 sẽ có:

\(\dfrac{2025-1}{2}+1=\dfrac{2024}{2}+1=1013\left(số\right)\)

Ta có: 1-3=5-7=...=2021-2023=-2

=>Sẽ có \(\dfrac{1013-1}{2}=\dfrac{1012}{2}=506\) cặp có tổng là -2 trong dãy số này

=>\(S=506\cdot\left(-2\right)+2025=2025-1012=1013\)

b: \(S=1+2-3-4+5+6-7-8+...+2021+2022-2023-2024\)

Từ 1 đến 2024 là: \(\dfrac{\left(2024-1\right)}{1}+1=2024\left(số\right)\)

Ta có: 1+2-3-4=5+6-7-8=...=2021+2022-2023-2024=-4

=>Sẽ có \(\dfrac{2024}{4}=506\) cặp có tổng là -4 trong dãy số này

=>\(S=506\cdot\left(-4\right)=-2024\)

Lời giải:
$a=1+5+5^2+5^3+...+5^{2022}+5^{2023}$

$5a=5+5^2+5^3+5^4+....+5^{2023}+5^{2024}$

$\Rightarrow 5a-a=5^{2024}-1$

$\Rightarrow 4a=5^{2024}-1$

$\Rightarrow 4a+1=5^{2024}\vdots 5^{2023}$ (đpcm)

1:

a: =23/27-11/17+4/27+28/17

=23/27+4/27+28/17-11/17

=1+1=2

b: \(=\dfrac{2}{3}\cdot\left(\dfrac{7}{9}+\dfrac{2}{9}\right)-\dfrac{2}{9}\)

=2/3-2/9

=6/9-2/9

=4/9

c: \(=\dfrac{11}{5}\cdot\dfrac{7}{3}-\dfrac{1}{3}\cdot\dfrac{11}{5}\)

=11/5(7/3-1/3)

=11/5*2

=22/5

d: \(=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{2024}{2023}=\dfrac{2024}{2}=1012\)

e: \(=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot...\cdot\dfrac{2022}{2023}=\dfrac{1}{2023}\)

thiếu rồi bạn ơi

`x` đâu ạ?

 A-B

A = 5⁰+5²+5⁴+...5²⁰²²

5²xA=5²+5⁴+...5²⁰²⁴ 

24xA = 5²⁰²⁴-1

A=\(\dfrac{5^{2024}-1}{24}\)

B = 5¹+5³+...5²⁰²³

B =5x(5⁰+5²+...5²⁰²²) = 5xA

M = A-B = A-5xA = -4A

M=\(\dfrac{1-5^{2024}}{6}\)

Vậy 24xA - 1 = 5²⁰²⁴

Nên 5²⁰²⁴ chia cho 3 dư 2 

( \(\dfrac{2}{123}\) + \(\dfrac{2023}{2022}\) )( \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) - \(\dfrac{2}{15}\))

=( \(\dfrac{2}{123}\) + \(\dfrac{2023}{2022}\) )( \(\dfrac{5}{15}\) - \(\dfrac{3}{15}\) - \(\dfrac{2}{15}\))

=( \(\dfrac{2}{123}\) + \(\dfrac{2023}{2022}\))( \(\dfrac{5-3-2}{15}\))

=( \(\dfrac{1}{123}\) + \(\dfrac{2023}{2022}\)). \(\dfrac{0}{15}\)

= ( \(\dfrac{1}{123}+\dfrac{2023}{2022}\)).0

= 0