Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, \(M=\frac{3}{2}\cdot\frac{4}{3}\cdot\cdot\cdot\cdot\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{3.4...2019}{2.3...2018}=\frac{2019}{2}\)
b, c cùng 1 câu phải k
ta có: \(A=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2017}-\frac{1}{2018}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2017}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2018}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{2018}-\left(1+\frac{1}{2}+...+\frac{1}{1009}\right)\)
\(=\frac{1}{1010}+\frac{1}{1011}+...+\frac{1}{2018}=B\)
\(\Rightarrow\frac{A}{B}=1\Rightarrow\left(\frac{A}{B}\right)^{2018}=1^{2018}=1\)
A,\(M=\frac{3}{2}\cdot\frac{4}{3}....\frac{2018}{2017}\cdot\frac{2019}{2018}=\frac{4\cdot3...2019}{2\cdot3...2018}=\frac{2019}{2}\)
NHA
HỌC TỐT
\(A=\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+\frac{6}{2}+..........+\frac{2019}{2}=\frac{3+4+5+..............+2019}{2}.\)
ta có 3+4+5+......+2019=(3+2019)2016:2=2038176
=>\(\frac{2038176}{2}=1019088\)
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{2017}\left(1+2+...+2017\right)\)
\(=1+\frac{1}{2}.\frac{2\left(2+1\right)}{2}+\frac{1}{3}.\frac{3\left(3+1\right)}{2}+....+\frac{1}{2017}.\frac{2017\left(2017+1\right)}{2}\)
\(=1+\frac{2.3}{2.2}+\frac{3.4}{3.2}+....+\frac{2017.2018}{2017.2}\)
\(=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+...+\frac{2018}{2}\)
\(=\frac{2+3+4+...+2018}{2}\)
\(=\frac{\frac{2018\left(2018+1\right)}{2}-1}{2}\)
\(=1018585\)
\(\left(\frac{1}{2}-1\right)\times\left(\frac{1}{3}-1\right)\times\left(\frac{1}{4}-1\right)\times...\times\left(\frac{1}{2016}-1\right)\times\left(\frac{1}{2017}-1\right)\)
\(=\left(-\frac{1}{2}\right)\times\left(-\frac{2}{3}\right)\times\left(-\frac{3}{4}\right)\times...\times\left(-\frac{2015}{2016}\right)\times\left(-\frac{2016}{2017}\right)\)
\(=\frac{1}{2017}\)
1/2017
Câu này dễ lắm, bn chỉ cần tính ra rùi chia mẫu cho tử, cái nào zống nhau thì bỏ nha
\(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)...\left(\frac{1}{2016}-1\right)\left(\frac{1}{2017}-1\right)\)
\(=\frac{-1}{2}.\frac{-2}{3}...\frac{-2015}{2016}.\frac{-2016}{2017}\)
\(=\frac{1.2...2015.2016}{2.3...2016.2017}\) ( tử số có chẵn số hạng )
\(=\frac{1}{2017}\)
\(P=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).....\left(\frac{1}{2017}-1\right)\left(\frac{1}{2018}-1\right)\)
\(P=\left(\frac{-1}{2}\right)\left(\frac{-2}{3}\right)\left(\frac{-3}{4}\right).....\left(\frac{-2016}{2017}\right)\left(\frac{-2017}{2018}\right)\)
\(P=\frac{\left(-1\right)\left(-2\right)\left(-3\right)\left(-4\right)....\left(-2017\right)}{2.3.4......2017.2018}\)
\(P=\frac{\left(-1\right)\left[\left(-2\right)\left(-3\right)\right]\left[\left(-4\right)\left(-5\right)\right]...\left[\left(-2016\right)\left(-2017\right)\right]}{\left[2.3\right]\left[4.5\right]....\left[2016.2017\right].2018}\)
\(P=\frac{\left(-1\right)\left[2.3\right]\left[4.5\right]....\left[2016.2017\right]}{\left[2.3\right]\left[4.5\right].....\left[2016.2017\right].2018}=\frac{-1}{2018}\)
B=1-1/2017
B=2016/2017
(Vì ta đưa bài toán về cách khử đi .VD:1/2+1/4+1/8=1-1/2+1/2-1/4+1/4-1/8 =1-1/2+1/2-1/4+1/4-1/8=1-1/8=7/8)
(Tui mới học lớp 5 thôi)
Giải
B=\(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)+...\left(1-\frac{1}{1+2+3+...+2017}\right)\)
=\(\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)...\left(1-\frac{1}{\left(1+2017\right).2017:2}\right)\)
=\(\frac{2}{3}.\frac{5}{6}...\frac{2018.2017:2-1}{2018.2017:2}\)
=\(\frac{4}{6}.\frac{10}{12}...\frac{\left(2018.2017:2-1\right).2}{2018.2017}\)
=\(\frac{1.4}{2.3}.\frac{2.5}{3.4}...\frac{2016.2019}{2017.2018}\)
=\(\frac{1.2...2016}{3.4....2018}.\frac{4.5...2019}{2.3...2017}\)
=\(\frac{2}{2017.2018}.\frac{2018.2019}{2.3}\)
=\(\frac{2019}{2017.3}=\frac{2019}{6051}=\frac{613}{2017}\)