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26 tháng 7 2015

=\(\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}...\cdot\frac{899}{900}=\frac{1.3}{2.2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}...\cdot\frac{29\cdot31}{30\cdot30}=\frac{1.2.3.4...29\cdot3.4.5...30.31}{2.2.3.3.4.4...30.30}=\frac{1.31}{2.30}=\frac{31}{60}\)

26 tháng 7 2015

\(A=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}....\frac{899}{30^2}=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right)...\left(29.31\right)}{\left(2.3.4...30\right).\left(2.3.4...30\right)}=\frac{\left(1.2....29\right).\left(3.4.5...31\right)}{\left(2.3.4...30\right).\left(2.3.4..30\right)}=\frac{1.31}{30.2}=\frac{31}{60}\)

NV
2 tháng 3 2022

\(1-\dfrac{1}{n^2}=\dfrac{n^2-1}{n^2}=\dfrac{\left(n-1\right)\left(n+1\right)}{n^2}\)

Do đó:

\(M=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\left(1-\dfrac{1}{4^2}\right)...\left(1-\dfrac{1}{30^2}\right)\)

\(=\dfrac{\left(2-1\right)\left(2+1\right)}{2^2}.\dfrac{\left(3-1\right)\left(3+1\right)}{3^2}.\dfrac{\left(4-1\right)\left(4+1\right)}{4^2}...\dfrac{\left(30-1\right)\left(30+1\right)}{30^2}\)

\(=\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.\dfrac{3.5}{4^2}...\dfrac{29.31}{30^2}=\dfrac{1.2.3...29}{2.3.4...30}.\dfrac{3.4.5...31}{2.3.4...30}\)

\(=\dfrac{1}{30}.\dfrac{31}{2}=\dfrac{31}{60}\)

15 tháng 8 2018

\(A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)...\left(1-\frac{1}{900}\right)\)

\(A=\frac{3}{4}\cdot\frac{8}{9}\cdot...\cdot\frac{899}{900}\)

\(A=\frac{\left(1\cdot3\right)\left(2\cdot4\right)...\left(29\cdot31\right)}{\left(2\cdot2\right)\left(3\cdot3\right)...\left(30\cdot30\right)}\)

\(A=\frac{\left(1\cdot2\cdot..\cdot29\right)\left(3\cdot4\cdot...\cdot31\right)}{\left(2\cdot3\cdot...\cdot30\right)\left(2\cdot3\cdot...\cdot30\right)}\)

\(A=\frac{1\cdot31}{30\cdot2}\)

\(A=\frac{31}{60}\)

20 tháng 5 2021

\(\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{900}-1\right)\)

\(=\left(\frac{1}{2}-1\right)\left(\frac{1}{2}+1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}-1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{30}-1\right)\left(\frac{1}{30}+1\right)\)

\(=\frac{-1}{2}.\frac{3}{2}.\frac{-2}{3}.\frac{4}{3}.\frac{-3}{4}.\frac{5}{4}...\frac{-29}{30}.\frac{31}{30}=-\frac{31}{60}\)

19 tháng 3 2022

(14−1)(19−1)(116−1)...(1900−1)

=(12−1)(12+1)(13−1)(13+1)(14−1)(14+1)...(130−1)(130+1)

27 tháng 3 2019

3. \(M=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{10.11.12}\)

\(\Leftrightarrow2M=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{10.11.12}\)

\(\Leftrightarrow2M=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{10.11}-\frac{1}{11.12}\)

\(\Leftrightarrow2M=\frac{1}{1.2}-\frac{1}{11.12}\)

\(\Leftrightarrow2M=\frac{1}{2}-\frac{1}{132}\)

\(\Leftrightarrow2M=\frac{65}{132}\)

\(\Leftrightarrow M=\frac{65}{132}\div2\)

\(\Leftrightarrow M=\frac{65}{264}\)

27 tháng 3 2019

1\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{899}{900}\)

\(\Leftrightarrow A=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{29.31}{30.30}\)

\(\Leftrightarrow A=\frac{1.3.2.4.3.5...29.31}{2.2.3.3.4.4...30.30}\)

\(\Leftrightarrow A=\frac{\left(1.2.3....29\right)\left(3.4.5...31\right)}{\left(2.3.4...30\right)\left(2.3.4...30\right)}\)

\(\Leftrightarrow A=\frac{1.31}{30.2}\)

\(\Leftrightarrow A=\frac{31}{60}\)

22 tháng 4 2021

Tìm y:

-y:1/2-5/2=4+1/2

-y:1/2 = 4+1/2+5/2

-y:1/2 = 7

-y = 7.2

y = -14

Vậy y = -14

31 tháng 3 2019

Đặt\(A=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}\)

\(\frac{1}{101}>\frac{1}{102}>\frac{1}{103}>...>\frac{1}{300}\)

\(\Rightarrow\left(\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+...+\frac{1}{300}\right)\)\(>\left(\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}\right)+\left(\frac{1}{300}+\frac{1}{300}+...+\frac{1}{300}\right)\)(mỗi cái trong ngoặc là một trăm phân số)

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+...+\frac{1}{300}>\left(\frac{1}{200}\right).100+\left(\frac{1}{300}\right).100\)

\(\Rightarrow A>\frac{1}{2}+\frac{1}{3}\)

\(\Rightarrow A>\frac{5}{6}\)

Mà 5/6>2/3=>A>2/3

\(\Rightarrow\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{300}\)

Đặt A = \(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{300}\)

Vì \(\frac{1}{101}>\frac{1}{102}>\frac{1}{103}>...>\frac{1}{300}\)

\(\Rightarrow\left(\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+....\frac{1}{200}\right)+\left(\frac{1}{201}+\frac{1}{202}+\frac{1}{103}+.....\frac{1}{300}\right)>\left(\frac{1}{200}+\frac{1}{200}+\frac{1}{200}\right)\)

Tự làm tiếp nhé !!!