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\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right)\) \(.....\left(1-\frac{1}{2014}\right)\)
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23 tháng 9 2017
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...........\left(1-\frac{1}{2014}\right)\)
\(=\left(\frac{2}{2}-\frac{1}{2}\right)\left(\frac{3}{3}-\frac{1}{3}\right).........\left(\frac{2014}{2014}-\frac{1}{2014}\right)\)
\(=\frac{1}{2}.\frac{2}{3}............\frac{2013}{2014}\)
\(=\frac{1}{2014}\)
14 tháng 12 2018
\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{n^2}\right)\)
\(=\left(\frac{2^2-1}{2^2}\right)\left(\frac{3^2-1}{3^2}\right)\left(\frac{4^2-1}{4^2}\right)...\left(\frac{n^2-1}{n^2}\right)\)
\(=\text{[}\frac{\left(2-1\right)\left(2+1\right)}{2^2}\text{]}.\text{[}\frac{\left(3-1\right)\left(3+1\right)}{3^2}\text{]}.\text{[}\frac{\left(4-1\right)\left(4+1\right)}{4^2}\text{]}...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)
\(=\left(\frac{1.3}{2^2}\right).\left(\frac{2.4}{3^2}\right).\left(\frac{3.5}{4^2}\right)...\text{[}\frac{\left(n-1\right)\left(n+1\right)}{n^2}\text{]}\)
\(=\frac{\text{[}1.2.3...\left(n-1\right)\text{]}.\text{[}3.4.5...\left(n+1\right)\text{]}}{\text{[}2.3.4...n\text{]}.\text{[}2.3.4...n\text{]}}\)
\(=\frac{1}{n}.\frac{n+1}{2}\)
\(=\frac{n+1}{2n}\)
9 tháng 5 2016
\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\)
\(=>B=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}=\frac{1\cdot2\cdot3}{2\cdot3\cdot4}=\frac{1}{4}\)
31 tháng 3 2020
\(\frac{1}{\left(x+1\right)\left(x+2\right)}-\frac{2}{\left(x+2\right)^2}+\frac{1}{\left(x+2\right)\left(x+3\right)}\)
\(=\frac{\left(x+2\right)\left(x+3\right)-2\left(x+1\right)\left(x+3\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
\(=\frac{\left(x+3\right)\left(x+2-2x-2\right)+x^2+2x+x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
\(=\frac{\left(x+3\right)\left(-x\right)+x^2+3x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
\(=\frac{-x^2-3x+x^2+3x+2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}=\frac{2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
1 tháng 4 2020
ĐKXD: x\(\ne\)-1,-2,-3
Ta có
\(\frac{1}{\left(x+1\right)\left(x+2\right)}\)-\(\frac{2}{\left(x+2\right)^2}\)+\(\frac{1}{\left(x+2\right)\left(x+3\right)}\)
=\(\frac{\left(x+2\right)\left(x+3\right)-2\left(x+1\right)\left(x+3\right)+\left(x+1\right)\left(x+2\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
=\(\frac{\left(x+2\right)\left(x+3+x+1\right)-2\left(x^2+4x+3\right)}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
=\(\frac{\left(x+2\right)\left(2x+4\right)-2x^2-8x-6}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
=\(\frac{2x^2+8x+8-2x^2-8x-6}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
=\(\frac{2}{\left(x+1\right)\left(x+2\right)^2\left(x+3\right)}\)
Chúc bạn học tốt
\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right).............\left(1-\dfrac{1}{2014}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right).\left(\dfrac{3}{3}-\dfrac{1}{3}\right)..............\left(\dfrac{2014}{2014}-\dfrac{1}{2014}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}................\dfrac{2013}{2014}\)
\(=\dfrac{1}{2014}\)
\(\left(1-\dfrac{1}{2}\right).\left(1-\dfrac{1}{3}\right)....\left(1-\dfrac{1}{2014}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right).\left(\dfrac{3}{3}-\dfrac{1}{3}\right).....\left(\dfrac{2014}{2014}-\dfrac{1}{2014}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}.....\dfrac{2013}{2014}\)
\(=\dfrac{1}{2014}\)
Chúc bạn học tốt!