Tính P=\(\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+....+2018}\right)\) giúp mk vs mn ơi
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11 tháng 6 2018
Cách 1 :
\(M=\left(7-\frac{1}{3}+\frac{3}{4}\right)-\left(6+\frac{2}{3}-\frac{1}{4}\right)\)
\(M=\left(\frac{84}{12}-\frac{4}{12}+\frac{9}{12}\right)-\left(\frac{72}{12}+\frac{8}{12}-\frac{3}{12}\right)\)
\(M=\frac{89}{12}-\frac{77}{12}\)
\(M=\frac{12}{12}\)
\(M=1\)
Cách 2 :
\(M=\left(7-\frac{1}{3}+\frac{3}{4}\right)-\left(6+\frac{2}{3}-\frac{1}{4}\right)\)
\(M=7-\frac{1}{3}+\frac{3}{4}-6-\frac{2}{3}+\frac{1}{4}\)
\(M=\left(7-6\right)-\left(\frac{2}{3}+\frac{1}{3}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)\)
\(M=1-1+1\)
\(M=1\)
11 tháng 6 2018
Cách 1:
\(M=\left(7-\frac{1}{3}+\frac{3}{4}\right)-\left(6+\frac{2}{3}-\frac{1}{4}\right)\)
\(M=\frac{89}{12}-\frac{77}{12}\)
\(M=1\)
Cách 2:
\(M=\left(7-\frac{1}{3}+\frac{3}{4}\right)-\left(6+\frac{2}{3}-\frac{1}{4}\right)\)
\(M=7-\frac{1}{3}+\frac{3}{4}-6-\frac{2}{3}+\frac{1}{4}\)
\(M=\left(7-6\right)-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{3}{4}+\frac{1}{4}\right)\)
\(M=1-1+1\)
\(M=1\)
1 tháng 12 2016
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+\frac{1}{4}\left(1+2+3+4\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\)
\(A=1+\frac{1+2}{2}+\frac{1+2+3}{3}+\frac{1+2+3+4}{4}+...+\frac{1+2+3+...+16}{16}\)
\(A=1+\frac{2\left(2+1\right):2}{2}+\frac{3\cdot\left(3+1\right):2}{3}+\frac{4\left(4+1\right):2}{4}+...+\frac{16\left(16+1\right):2}{16}\)
\(A=1+\frac{2+1}{2}+\frac{3+1}{2}+\frac{4+1}{2}+...+\frac{16+1}{2}\)
\(A=\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{17}{2}\)
\(A=\frac{2+3+4+5+...+17}{2}\)
\(A=\frac{152}{2}\)
\(A=76\)
A
20 tháng 6 2018
\(\Leftrightarrow2.\left(\frac{-1}{2}\right).\left(\frac{2}{3}\right)^2-3\left(-\frac{1}{3}\right)^2.\frac{2}{9}:x=3.\left(-\frac{1}{2}\right)-\frac{2}{3}\)
\(\Leftrightarrow-\frac{4}{9}-\frac{1}{3}.\frac{2}{9}:x=-\frac{3}{2}-\frac{2}{3}\)
\(\Leftrightarrow-\frac{4}{6}-\frac{2}{27}:x=-\frac{13}{6}\)
\(\Leftrightarrow\frac{2}{27}:x=-\frac{4}{9}:\frac{-13}{6}\)
\(\Leftrightarrow\frac{2}{27}:x=\frac{31}{18}\)
\(\Leftrightarrow x=\frac{2}{27}:\frac{31}{18}\)
\(\Rightarrow x=\frac{4}{93}\)
Vậy \(x=\frac{4}{93}\)
12 tháng 4 2019
cho 3 k
\(\left(1-\frac{1}{2^2}\right)\cdot\left(1-\frac{1}{3^2}\right)...\left(1-\frac{1}{10^2}\right)\)
=> \(\left(1-\frac{1}{2}\right)\left(1+\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1+\frac{1}{3}\right)\)\(...\left(1-\frac{1}{10}\right)\cdot\left(1+\frac{1}{10}\right)\)
=> \(\left(1-\frac{1}{2}\right)\cdot\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\cdot\cdot\frac{9}{10}\cdot\frac{10}{11}\)
=> \(\frac{1}{2}\cdot\frac{3\cdot2\cdot4\cdot\cdot\cdot9\cdot10}{2\cdot3\cdot3\cdot\cdot\cdot10\cdot11}=\frac{1}{2}\cdot\frac{11}{10}=\frac{11}{20}\)
Chúc bn học tốt !
12 tháng 4 2019
cho mk 3 k nha bn
thanks nhìu
bài này mk ko copy, ko chép mạng, tự nghĩ mất 6 phút .
có công thức rùi nha !
chúc bn học tốt
27 tháng 9 2020
\(B=1+\frac{1}{2}.\frac{\left(1+2\right).2}{2}+\frac{1}{3}.\frac{\left(1+3\right).3}{2}+...+\frac{1}{2018}.\frac{\left(1+2018\right).2018}{2}\)
\(=1+\frac{3}{2}+\frac{4}{2}+...+\frac{2019}{2}=1+\frac{3+4+...+2019}{2}=1+\frac{\left(3+2019\right)2017}{2}=2039188\)
TP
13 tháng 8 2017
Ta có:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)........\left(1-\frac{1}{2017}\right).\left(1-\frac{1}{2018}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.......\frac{2016}{2017}.\frac{2017}{2018}\)
Đởn giản hết sẽ còn là:
\(\Rightarrow B=\frac{1}{2018}\)
Ta có:
\(1-\frac{1}{1+2+...+k}=1-\frac{1}{\frac{k\left(k+1\right)}{2}}=\frac{k\left(k+1\right)-2}{k\left(k+1\right)}=\frac{\left(k-1\right)\left(k+2\right)}{k\left(k+1\right)}\)
Áp dụng biểu thức trên ta được:
\(P=\left(1-\frac{1}{1+2}\right)\left(1-\frac{1}{1+2+3}\right)...\left(1-\frac{1}{1+2+3+...+2018}\right)\)
\(P=\frac{1.4}{2.3}.\frac{2.5}{3.4}.\frac{3.6}{4.5}.....\frac{2017.2020}{2018.2019}\)
\(P=\frac{1}{2018}.\frac{2020}{3}=\frac{1010}{3027}\)