Tính B = \(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{2007}\right)\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2007^2}\right)\)
\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}......\frac{2007^2-1}{2007^2}\)
\(=\frac{\left(2-1\right)\left(2+1\right)}{2.2}.\frac{\left(3-1\right)\left(3+1\right)}{3.3}.\frac{\left(4-1\right)\left(4+1\right)}{4.4}....\frac{\left(2007-1\right)\left(2007+1\right)}{2007.2007}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{2006.2008}{2007.2007}\)
\(=\frac{\left(1.2.3.....2006\right)\left(3.4.5....2008\right)}{\left(2.3.4...2007\right)\left(2.3.4.....2007\right)}\)
\(=\frac{1.2008}{2007.2}=\frac{1004}{2007}\)
A/ \(\left(10\frac{3}{4}+3\frac{4}{5}\right)-\left(5\frac{3}{4}-1\frac{1}{5}\right)\)
\(=\left(10\frac{3}{4}-5\frac{3}{4}\right)+\left(3\frac{4}{5}+1\frac{1}{5}\right)\)
\(=5+5\)
\(=10\)
chúc bạn học tốt nha
B = -1/2 . (-2/3) . ....... . (-2006/2007)
= 1/2 . 2/3 . ....... . 2006/2007 ( vì số phân số là số chẵn )
= 1.2.3. ..... .2006/2.3.4. ...... .2007
= 1/2007
Tk mk nha
Ta có :
\(B=\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{2007}\right)\)
\(B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{2006}{2007}\)
\(B=\frac{1.2.3.....2006}{2.3.4.....2007}\)
\(B=\frac{1}{2007}\)
Vậy \(B=\frac{1}{2007}\)
Chúc bạn học tốt ~
Với mọi n thuộc N* ta có :
\(n^4+\frac{1}{4}=\left(n^4+2.\frac{1}{2}.n^2+\frac{1}{4}\right)-n^2=\left(n^2+\frac{1}{2}\right)^2-n^2\)
\(=\left(n^2+n+\frac{1}{2}\right)\left(n^2-n+\frac{1}{2}\right)\)
\(\Rightarrow N=\frac{\left(2^2+2+\frac{1}{2}\right)\left(2^2-2+\frac{1}{2}\right)...\left(2008^2+2008+\frac{1}{2}\right)\left(2008^2-2008+\frac{1}{2}\right)}{\left(1^2+1+\frac{1}{2}\right)\left(1^2-1+\frac{1}{2}\right)...\left(2007^2+2007+\frac{1}{2}\right)\left(2007^2-2007+\frac{1}{2}\right)}\)
\(=\frac{\left(2.3+\frac{1}{2}\right)\left(1.2+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right)...\left(2008.2009+\frac{1}{2}\right)}{\frac{1}{2}\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)...\left(2007.2008+\frac{1}{2}\right)}\)
\(=\frac{2008.2009+\frac{1}{2}}{\frac{1}{2}}=8068145\)
a)S=1+(-1/7)^1+(-1/7)^2+...+(-1/7)^2007
=>7S=7+(-1/7)^1+(1/7)^2+...+(-1/7)^2006
=>(7-1)S=6-(1/7)^2007
=>S=1-(-1/7^2007/6)
\(x\)là dấu nhân hả bạn? Nếu vậy thì mk làm cho nhé
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.......\cdot\frac{17}{18}\cdot\frac{18}{19}\cdot\frac{19}{20}=\frac{1}{20}\)
Vậy \(A=\frac{1}{20}\)
\(B=1\frac{1}{2}\cdot1\frac{1}{3}\cdot1\frac{1}{4}\cdot........\cdot1\frac{1}{2005}\cdot1\frac{1}{2006}\cdot1\frac{1}{2007}\)
\(B=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot......\cdot\frac{2006}{2005}\cdot\frac{2007}{2006}\cdot\frac{2008}{2007}=\frac{2008}{2}=1004\)
Vậy \(B=1004\)
DẤU CHẤM LÀ DẤU NHÂN
a,
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{19}{20}=\frac{1}{20}\)
b, \(1\frac{1}{2}.1\frac{1}{3}....1\frac{1}{2017}=\frac{3}{2}.\frac{4}{3}....\frac{2018}{2017}=\frac{2018}{2}=1009\)
\(A=\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x...x\frac{2006}{2007}=\frac{1}{2007}\)
k nha bạn