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15 tháng 12 2022

\(=1-\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\)

\(=1-\dfrac{6}{21}=\dfrac{15}{21}=\dfrac{5}{7}\)

14 tháng 1 2023

`2/[3.5]+2/[5.7]+2/[7.9]+....+2/[19.21]`

`=1/3-1/5+1/5-1/7+1/7-1/9+....+1/19-1/21`

`=1/3-1/21`

`=6/21`

8 tháng 5 2022

SOS

8 tháng 5 2022

hép☹

Giải:

\(B=\dfrac{3}{3\times5}+\dfrac{3}{5\times7}+\dfrac{3}{7\times9}+...+\dfrac{3}{48\times50}\) 

\(B=\dfrac{3}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{48\times50}\right)\) 

\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\) 

\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{50}\right)\)

\(B=\dfrac{3}{2}\times\dfrac{47}{150}\) 

\(B=\dfrac{47}{100}\) 

Chúc em học tốt!

20 tháng 9 2023

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\)

\(=2\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+\dfrac{1}{9\times11}\right)\)

\(=2\times\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\)

\(=1-\dfrac{1}{11}\)

\(=\dfrac{11}{11}-\dfrac{1}{11}\)

\(=\dfrac{10}{11}\)

HQ
Hà Quang Minh
Giáo viên
20 tháng 9 2023

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\\ =1-\dfrac{1}{11}\\ =\dfrac{10}{11}\)

3 tháng 5 2022

a) \(\left(1-\dfrac{1}{3}\right)\times\left(1-\dfrac{2}{5}\right)\times\left(1-\dfrac{2}{7}\right)\times\left(1-\dfrac{2}{9}\right)\)

\(=\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\times\left(\dfrac{5}{5}-\dfrac{2}{5}\right)\times\left(\dfrac{7}{7}-\dfrac{2}{7}\right)\times\left(\dfrac{9}{9}-\dfrac{2}{9}\right)\)

\(=\dfrac{2}{3}\times\dfrac{3}{5}\times\dfrac{5}{7}\times\dfrac{7}{9}\)

\(=\dfrac{2\times3\times5\times7}{3\times5\times7\times9}\)

\(=\dfrac{2}{9}\)

b) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}\)

\(=1-\dfrac{1}{9}\)

\(=\dfrac{9}{9}-\dfrac{1}{9}\)

\(=\dfrac{8}{9}\)

3 tháng 5 2022

Sửa câu b)

b) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}\)

 

Đặt \(A=\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}\)

\(2A=\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}\)

\(2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}\)

\(2A=1-\dfrac{1}{9}\)

\(2A=\dfrac{9}{9}-\dfrac{1}{9}\)

\(2A=\dfrac{8}{9}\)

\(A=\dfrac{8}{9}:2\)

\(A=\dfrac{8}{18}\)

\(A=\dfrac{4}{9}\)

Vậy : \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}=\dfrac{4}{9}\)

8 tháng 8 2023

`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`

8 tháng 8 2023

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

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

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

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

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

 

11 tháng 5 2015

\(\frac{2}{1\times3}+\frac{2}{3\times5}+...+\frac{2}{19\times21}=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{21}\)

                                                    \(=1-\frac{1}{21}=\frac{20}{21}\)

đúng cái nhé

22 tháng 10 2023

\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{13\times15}+\dfrac{2}{15\times17}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{17}\)

\(=1-\dfrac{1}{17}\)

\(=\dfrac{16}{17}\)

22 tháng 10 2023

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{15\cdot17}\)

\(=2-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{17}\)

\(=2-\dfrac{1}{17}\)

\(=\dfrac{33}{17}\)

27 tháng 4 2017

a)\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{99.101}=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)

b)ĐK: \(n\ne-5\)

\(A=\dfrac{n-2}{n+5}=\dfrac{n+5-7}{n+5}=1-\dfrac{7}{n+5}\)

Để A nguyên thì \(\dfrac{n-2}{n+5}\)phải nguyên <=> \(\dfrac{7}{n+5}\) nguyên mà n là số nguyên <=> 7 chia hết cho n+5 hay n+5 là Ư(7)

Mà Ư(7)={-1;1;-7;7}

Ta có bảng sau:

n+5 -1 1 -7 7
n -6(TM) -4(TM) -12(TM) 2(TM)

Vậy n={-6;-4;-12;2} thì A nguyên

27 tháng 4 2017

a. \(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)

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

\(=1-\dfrac{1}{100}\)

\(=\dfrac{99}{100}\)

b, Ta có: \(A=\dfrac{n-2}{n+5}=\dfrac{n+5-7}{n+5}=1-\dfrac{7}{n+5}\)

Để \(A\in Z\) thì \(\dfrac{n-2}{n+5}\in Z\Rightarrow7⋮n+5\Leftrightarrow n+5\in U\left(7\right)=\left\{\pm1;\pm7\right\}\)

Lập bảng giá trị:

\(n+5\) \(1\) \(-1\) \(7\) \(-7\)
\(n\) \(-4\) \(-6\) \(2\) \(-12\)

Vậy, với \(x\in\left\{-12;-6;-4;2\right\}\) thì \(A=\dfrac{n-2}{n+5}\in Z\)