A=(1-\(\dfrac{1}{2^2}\)).(1-\(\dfrac{1}{3^2}\)).............(1-\(\dfrac{1}{10^2}\)) help me
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
21 tháng 8 2023
\(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=1\dfrac{1989}{1991}\)
\(\Rightarrow2\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3980}{1991}\)
\(\Rightarrow2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3980}{1991}\)
\(\Rightarrow2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)
\(\Rightarrow2\left(1-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)
\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{3980}{1991}.\dfrac{1}{2}\)
\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{1990}{1991}\)
\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{1990}{1991}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{1991}\)
\(\Rightarrow x+1=1991\)
\(\Rightarrow x=1990\)
BD
20 tháng 8 2023
Sửa đề
\(\dfrac{2}{1^2}\cdot\dfrac{6}{2^2}\cdot\dfrac{12}{3^3}\cdot.......\cdot\dfrac{110}{10^2}\cdot x=-20\)
\(\dfrac{2}{1\cdot1}\cdot\dfrac{2\cdot3}{2\cdot2}\cdot\cdot\cdot\cdot\dfrac{11\cdot10}{10\cdot10}\cdot x=-20\)
\(\dfrac{\left(2\cdot3\cdot4\cdot....\cdot11\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot\dfrac{\left(1\cdot2\cdot3\cdot4\cdot5\cdot...\cdot10\right)}{\left(1\cdot2\cdot3\cdot4\cdot...\cdot10\right)}\cdot x=-20\)
\(11\cdot x=-20\\ x=-\dfrac{20}{11}\)
NT
2
15 tháng 4 2017
Ta có :
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+..............+\dfrac{1}{2^{10}}\)
\(2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+.................+\dfrac{1}{2^9}\)
\(\Rightarrow2A-A=\left(\dfrac{1}{2}+...........+\dfrac{1}{2^{10}}\right)-\left(1+\dfrac{1}{2}+..........+\dfrac{1}{2^9}\right)\)
\(\Rightarrow A=1-\dfrac{1}{2^{10}}\)
\(A=\dfrac{2^{10}-1}{2^{10}}\)
~ Chúc bn học tốt ~
15 tháng 4 2017
A=\(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\)
2A=2(\(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\))
2A= \(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^9}\)
2A-A=1-\(\dfrac{1}{2^{10}}\)=\(\dfrac{1023}{1024}\)
Q
1 tháng 4 2017
Đặt vế đầu là A, vế sau là B.
Vế A:
- Tử:
\(\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}\)
\(=100\left(\dfrac{1}{99}+\dfrac{1}{98}+\dfrac{1}{97}+...+\dfrac{1}{3}+\dfrac{1}{2}+\dfrac{1}{100}\right)\)
\(=100\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{98}+\dfrac{1}{99}+\dfrac{1}{100}\right)\)
Vậy:
\(A=\dfrac{\dfrac{1}{99}+\dfrac{2}{98}+\dfrac{3}{97}+...+\dfrac{99}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}\\ =\dfrac{50\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+..+\dfrac{1}{100}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}}\\ \Rightarrow A=50\)
Vế B:
- Tử:
\(92-\dfrac{1}{9}-\dfrac{1}{10}-...-\dfrac{92}{100}\\ =\left(1-\dfrac{1}{9}\right)+\left(1-\dfrac{2}{10}\right)+...+\left(1-\dfrac{92}{100}\right)\\ =\dfrac{8}{9}+\dfrac{8}{10}+...+\dfrac{8}{100}\\ =\dfrac{40}{45}+\dfrac{40}{50}+...+\dfrac{40}{500}\\ =40\left(\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}\right)\)
Vậy:
\(B=\dfrac{92-\dfrac{1}{9}-\dfrac{1}{10}-...-\dfrac{92}{100}}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}}\\ =\dfrac{40\left(\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{55}+...+\dfrac{1}{500}\right)}{\dfrac{1}{45}+\dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{500}}\\ \Rightarrow B=40\)
Từ 2 vế trên ta tính được \(\dfrac{A}{B}=\dfrac{50}{40}=\dfrac{5}{4}\)
AH
Akai Haruma
Giáo viên
12 tháng 9 2018
Lời giải:
Xét \(1+\frac{1}{n^2}+\frac{1}{(n+1)^2}=\frac{n^2+1}{n^2}+\frac{1}{(n+1)^2}\)
\(=\frac{(n+1)^2-2n}{n^2}+\frac{1}{(n+1)^2}=\left(\frac{n+1}{n}\right)^2+\frac{1}{(n+1)^2}-\frac{2}{n}\)
\(=\left(\frac{n+1}{n}-\frac{1}{n+1}\right)^2=\left(1+\frac{1}{n}-\frac{1}{n+1}\right)^2\)
\(\Rightarrow \sqrt{1+\frac{1}{n^2}+\frac{1}{(n+1)^2}}=1+\frac{1}{n}-\frac{1}{n+1}\)
Áp dụng vào bài toán suy ra:
\(A=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}-\frac{1}{4}+...+1+\frac{1}{2017}-\frac{1}{2018}\)
\(=2016+\frac{1}{2}-\frac{1}{2018}=2016,5-\frac{1}{2018}\)
Y
27 tháng 6 2023
\(a,\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{a-9}\left(dkxd:a\ne9,a\ge0\right)\)
\(=\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)-3\left(\sqrt{a}-3\right)-a+2}{a-9}\)
\(=\dfrac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{a-9}\)
\(=\dfrac{11}{a-9}\)
\(b,\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\left(dkxd:x\ge0,x\ne1\right)\)
\(=\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)
\(=\dfrac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)-\left(x+\sqrt{x}+1\right)}{x\sqrt{x}-1}\)
\(=\dfrac{x+2+x-1-x-\sqrt{x}-1}{x\sqrt{x}-1}\)
\(=\dfrac{x-\sqrt{x}}{x\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(x+\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\\ =\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)
27 tháng 6 2023
bạn ơi có phải \(x\sqrt{x}\) là \(\left(\sqrt{x}\right)^3\) đúng ko ạ
AL
18 tháng 4 2021
B=\(\dfrac{1}{2}:\left(-1\dfrac{1}{2}\right):1\dfrac{1}{3}:....:\left(-1\dfrac{1}{100}\right)\)
=\(\dfrac{1}{2}:\dfrac{-3}{2}:\dfrac{4}{3}:....:\dfrac{-101}{100}\)
=\(\dfrac{1}{2}.\dfrac{-2}{3}.\dfrac{3}{4}........\dfrac{-100}{101}\)
=\(\dfrac{1.\left(-2\right).3......\left(-100\right)}{2.3.4...........101}\)
=\(\dfrac{1}{101}\)
23 tháng 1 2022
\(=\left(\dfrac{2007}{2}+1\right)+\left(\dfrac{2006}{3}+1\right)+...+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)+1\)
\(=\dfrac{2009}{2}+\dfrac{2009}{3}+...+\dfrac{2009}{2008}+\dfrac{2009}{2009}\)
\(=2009\left(\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2009}\right)\)
