\(S=\left(\frac{1}{2}-1\right)\div\left(\frac{1}{3}-1\right)\div\left(\frac{1}{4}-1\right)\div...\div\left(\frac{1}{50}-1\right)\)
vậy S bằng bao nhiêu ?
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
PA
8 tháng 9 2016
\(\left[\left(1+\frac{1}{x^2}\right)\div\left(1+2x+x^2\right)+\frac{2}{\left(x+1\right)^3}\times\left(1+\frac{1}{x}\right)\right]\div\frac{x-1}{x^3}\)
\(=\left[\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{2}{\left(x+1\right)^3}\times\frac{x+1}{x}\right]\div\frac{x-1}{x^3}\)
\(=\left(\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{1}{\left(x+1\right)^2}\times\frac{2}{x}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\left(\frac{x^2+1}{x^2}+\frac{2}{x}\right)\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x^3+2x^2+x}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x^2+2x+1\right)}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x+1\right)^2}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\frac{1}{x^2}\times\frac{x^3}{x-1}\)
\(=\frac{x}{x-1}\)
KN
25 tháng 8 2020
\(4.\left(\frac{1}{4}\right)^2+25\left[\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\right]:\left(\frac{3}{2}\right)^3=4.\frac{1}{16}+25\left(\frac{27}{64}.\frac{64}{125}\right).\frac{8}{27}\)
\(=\frac{1}{4}+25.\frac{27}{125}.\frac{8}{27}=\frac{1}{4}+\frac{8}{5}=\frac{37}{20}\)
\(2^3+3\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\frac{1}{2}\right]-8=8+3-1+4.2-8=10\)
22 tháng 6 2016
=\(2\left(\frac{1}{2}-\frac{1}{2.3}\right).2\left(\frac{1}{2}-\frac{1}{3.4}\right)....2\left(\frac{1}{2}-\frac{1}{99.100}\right)\)
=\(2^{89}\left(\frac{1}{2}.98-\frac{1}{2}+\frac{1}{100}\right)\)
\(=2^{98}.\left(49-\frac{49}{100}\right)=\frac{2^{98}.4851}{100}\)
2 tháng 4 2017
B=\(\left(1-\dfrac{1}{1+2}\right)\). \(\left(1-\dfrac{1}{1+2+3}\right)\).....\(\left(1-\dfrac{1}{1+2+...+100}\right)\)
B=\(\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot...\cdot\left(1-\dfrac{1}{\left(1+100\right)\cdot100:2}\right)\)
B=\(\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot...\cdot\dfrac{101\cdot100:2-1}{101\cdot100:2}\)
B=\(\dfrac{4}{6}\cdot\dfrac{10}{12}\cdot...\cdot\dfrac{\left(101.100:2-1\right).2}{101.100}\)
B=\(\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}\cdot...\cdot\dfrac{99.102}{100.101}\)
B=\(\dfrac{1.2.3.4.....99}{3.4.5....100}.\dfrac{4.5.6.....102}{3.4.5.....101}\)
B=\(\dfrac{2}{100}\).\(\dfrac{102}{3}\)
B=\(\dfrac{17}{25}\)
15 tháng 1 2017
Ta có công thức :
\(1-\frac{1}{k^2}=\frac{k^2-1^2}{k^2}=\frac{\left(k+1\right)\left(k-1\right)}{k^2}\)
Áp dụng công thức trên ta được :
\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)
\(=\frac{2^2-1^2}{2^2}.\frac{3^2-1^2}{3^2}.\frac{4^2-1^2}{4^2}....\frac{n^2-1^2}{n^2}\)
\(=\frac{\left(2+1\right)\left(2-1\right)}{2.2}.\frac{\left(3+1\right)\left(3-1\right)}{3.3}.\frac{\left(4+1\right)\left(4-1\right)}{4.4}...\frac{\left(n+1\right)\left(n-1\right)}{n.n}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{\left(n+1\right)\left(n-1\right)}{n.n}\)
\(=\frac{\left[1.2.3.....\left(n+1\right)\right].\left[3.4.5...\left(n-1\right)\right]}{\left(2.3.4....n\right)\left(2.3.4....n\right)}\)
\(=\left(n+1\right).\frac{1}{2n}=\frac{n+1}{2n}\)
15 tháng 1 2017
mới lớp 6 mà giải đc toan lớp 8 , anh đây thuông minh quá ))
14 tháng 10 2016
Bấm máy tính:
E = \(\frac{4}{3}+\frac{1}{4}+\frac{3}{5}:\frac{4}{5}\)
E = \(\frac{4}{3}+\frac{1}{4}+\frac{3}{4}\)
E = \(\frac{7}{3}\)
Vậy E = \(\frac{7}{3}\)
Ta có S = ( 1/2 - 1) : ( 1/3 - 1) : (1/4 - 1) :... : ( 1/50 - 1)
S = -1/2 : ( -2/3) : ( -3/4) : ... : ( -49/ 50)
S= -1/2 x (-3/2) x ( -4/3) x ... x (-50/49)
S= -1/2 x 1/3 x 50
S= -25/3
cho mình thêm câu trả lời đi