\(\frac{1}{200}-\left(\frac{1}{200.199}-\frac{1}{199.198}-...-\frac{1}{3.2}-\frac{1}{2.1}\right)\)]
Giúp mình với
\(\frac{1}{200}-\left(\frac{1}{200.199}-\frac{1}{199.198}-...-\frac{1}{3.2}-\frac{1}{2.1}\right)\)
giúp mình với mai mình đi hc rồi
link nè bạn
/hoi-dap/question/88514.html
hoặc bạn sang trang 3 của hỏi đáp toán hoc24 sẽ thấy nhé
Bài 1 : Tính nhanh:
a) \(\left(\frac{5}{19}-\frac{1}{511}+\frac{7}{12}\right)-\left(-\frac{1}{511}-\frac{1}{2}+\frac{5}{19}\right)\)
b) \(-\left(\frac{13}{25}-\frac{4}{191}+\frac{2}{51}\right)+\left(\frac{4}{191}+\frac{2}{51}+\frac{3}{5}\right)\)
c) \(\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
Bài 1:
a) \(\left(\frac{5}{19}-\frac{1}{511}+\frac{7}{12}\right)-\left(-\frac{1}{511}-\frac{1}{2}+\frac{5}{19}\right)\)
= \(\frac{5}{19}-\frac{1}{511}+\frac{7}{12}+\frac{1}{511}+\frac{1}{2}-\frac{5}{19}\)
= \(\left(\frac{5}{19}-\frac{5}{19}\right)+\left(\frac{1}{511}-\frac{1}{511}\right)+\left(\frac{7}{12}+\frac{1}{2}\right)\)
= 0 + 0 + \(\frac{13}{12}\)
= \(\frac{13}{12}\).
b) \(-\left(\frac{13}{25}-\frac{4}{191}+\frac{2}{51}\right)+\left(\frac{4}{191}+\frac{2}{51}+\frac{3}{5}\right)\)
= \(-\frac{13}{25}+\frac{4}{191}-\frac{2}{51}+\frac{4}{191}+\frac{2}{51}+\frac{3}{5}\)
= \(\left(-\frac{13}{25}+\frac{3}{5}\right)+\left(\frac{4}{191}+\frac{4}{191}\right)+\left(\frac{2}{51}-\frac{2}{51}\right)\)
= \(\frac{2}{25}+\frac{8}{191}+0\)
= \(\frac{582}{4775}\).
Mình chỉ làm câu a) và câu b) thôi nhé.
Chúc bạn học tốt!
Tính hợp lí
1/200- 1/200.199- 1/199.198- .....-1/3.2- 1/2.1
\(\frac{1}{200}-\frac{1}{200.199}-\frac{1}{199.198}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{200}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{198.199}+\frac{1}{199.200}\right)\)
\(=\frac{1}{200}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{198}-\frac{1}{199}+\frac{1}{199}-\frac{1}{200}\right)\)
\(=\frac{1}{200}-\left(1+\frac{1}{200}\right)\)
\(=\left(\frac{1}{200}-\frac{1}{200}\right)-1\)
\(=0-1\)
\(=-1\)
Tính hợp lí
1/200- 1/200.199-1/199.198-.....-1/3.2-1/2.1
a) \(A=\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-\frac{1}{197.196}-....-\frac{1}{3.2}-\frac{1}{2.1}\)
b) \(B=1-\frac{2}{3.5}-\frac{2}{5.7}-\frac{2}{7.9}-....-\frac{2}{61.63}-\frac{2}{63-65}\)
\(a)\) \(A=\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-\frac{1}{197.196}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(A=\frac{1}{199}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{197.198}+\frac{1}{198.199}\right)\)
\(A=\frac{1}{199}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{197}-\frac{1}{198}+\frac{1}{198}-\frac{1}{199}\right)\)
\(A=\frac{1}{199}-\left(1-\frac{1}{199}\right)\)
\(A=\frac{1}{199}-1+\frac{1}{199}\)
\(A=\frac{-197}{199}\)
Chúc bạn học tốt ~
a) \(A=\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-\frac{1}{197.196}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(A=\frac{1}{199}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...\frac{1}{197.198}+\frac{1}{198.199}\right)\)
\(A=\frac{1}{199}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{197}-\frac{1}{198}+\frac{1}{198}-\frac{1}{199}\right)\)
\(A=\frac{1}{199}-(1-\frac{1}{199})\)
\(A=\frac{1}{199}-1+\frac{1}{199}\)
\(A=\frac{-197}{199}\)
Vậy \(A=\frac{-197}{199}\)
b) cũng thế
mọi người giúp mình với: A=\(\frac{1}{2003.2002}-\frac{1}{2002.2001}-...\frac{1}{3.2}-\frac{1}{2.1}\)
Ta có :
\(A=\frac{1}{2003\cdot2002}-\frac{1}{2002\cdot2001}-...-\frac{1}{3\cdot2}-\frac{1}{2\cdot1}\)
\(A=-\left(\frac{1}{2003\cdot2002}+\frac{1}{2002\cdot2001}+...+\frac{1}{3\cdot2}+\frac{1}{2\cdot1}\right)\)
\(A=-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{2001\cdot2002}+\frac{1}{2002\cdot2003}\right)\)
\(A=-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2002}-\frac{1}{2003}\right)\)
\(A=-\left(1-\frac{1}{2003}\right)\)
\(A=-\frac{2002}{2003}\)
\(A=\frac{1}{2003.2002}-\frac{1}{2002.2001}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2001.2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2001}-\frac{1}{2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)
\(=-\left(1-\frac{1}{2002}\right)+\frac{1}{2002}.\frac{1}{2003}\)
\(=-1+\frac{1}{2002}.+\frac{1}{2002}.\frac{1}{2003}\)
\(=-1+\frac{1}{2002}\left(1+\frac{1}{2003}\right)\)
\(=-1+\frac{1}{2002}.\frac{2004}{2003}\)
\(=-1+\frac{2}{2003}\)
\(=\frac{-2003+2}{2003}\)
\(=\frac{-2001}{2003}\)
cho C=\(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-.......-\frac{1}{3.2}-\frac{1}{2.1}\)
giúp với ạ mình đang cần gấp
\(C=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(C=\frac{1}{100}-\left(\frac{1}{1}-\frac{1}{100}\right)=\frac{1}{100}-\frac{99}{100}=\frac{-98}{100}=\frac{-49}{50}\)
Tính nhanh:
a)\(A=\frac{1}{2015}-\frac{1}{2015.2014}-\frac{1}{2014.2013}-...-\)\(\frac{1}{3.2}-\frac{1}{2.1}\)
b)\(B=54-\frac{1}{2}.\left(1+2\right)-\frac{1}{3}.\left(1+2+3\right)\)\(-...-\frac{1}{12}.\left(1+2+3+..+12\right)\)
Giải Hộ Mình Pls
A=1/2015-1/2015.2014-....-1/3.2-1/2.1
A=1/2015-[1/2015.2014+1/2014.2013+....+1/3.2+1/2.1]
A=1/2015-[1/1.2+1/2.3+....1/2014.2015]
A=1/2015-[1-1/2+1/2-1/3+...+1/2014-1/2015]
A=1/2015-[1-2015]
A=1/2015-1+1/2015
A=[1/2015+1/2015]-1
A=2/2015-1
A=-2013/2015
Bài 1: Tính
a)\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{n+1}\right)\)
b)\(\frac{1}{2000.1999}-\frac{1}{1999-1998}-\frac{1}{1998-1997}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
c)\(-3+\frac{1}{1+\frac{1}{3+\frac{1}{1+\frac{1}{3}}}}\)
a)
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)...\left(1-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{n}{n+1}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot n}{2\cdot3\cdot4\cdot...\cdot\left(n+1\right)}\)
\(=\frac{1}{n+1}\)
a)\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right).....\left(1-\frac{1}{n+1}\right)\)
\(=\left(\frac{2}{2}-\frac{1}{2}\right)\left(\frac{3}{3}-\frac{1}{3}\right)\left(\frac{4}{4}-\frac{1}{4}\right)....\left(\frac{n+1}{n+1}-\frac{1}{n+1}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.....\frac{n}{n+1}=\frac{1}{n+1}\)