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MM
1
H9
HT.Phong (9A5)
CTVHS
8 tháng 10 2023
\(A=\left(2020\times2019+2019\times2018\right)\times\left(1+\dfrac{1}{2}:1\dfrac{1}{2}-1\dfrac{1}{3}\right)\)
\(A=\left[2019\times\left(2020+2018\right)\right]\times\left(1+\dfrac{1}{2}:\dfrac{3}{2}-\dfrac{4}{3}\right)\)
\(A=4038\times2019\times\left(1+\dfrac{1}{3}-\dfrac{4}{3}\right)\)
\(A=4038\times2019\times0\)
\(A=0\)
NM
20 tháng 9 2023
\(x\times2+x\times\dfrac{1}{5}=1.\dfrac{1}{3}\\ \Rightarrow x\times\left(2+\dfrac{1}{5}\right)=\dfrac{1}{3}\\ \Rightarrow x\times\dfrac{11}{5}=\dfrac{1}{3}\\ \Rightarrow x=\dfrac{5}{33}\)
TD
4
30 tháng 6 2018
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times...\times\left(1-\frac{1}{2018}\right)\times\left(1-\frac{1}{2019}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{2017}{2018}\times\frac{2018}{2019}\)
\(=\frac{1\times2\times3\times...\times2017\times2018}{2\times3\times4\times...\times2018\times2019}\)
\(=\frac{1}{2019}\)
15 tháng 5 2022
\(F=1\dfrac{1}{5}\times1\dfrac{1}{6}\times1\dfrac{1}{7}\times\cdot\cdot\cdot\times1\dfrac{1}{2019}\times1\dfrac{1}{2020}\)
\(F=\dfrac{6}{5}\times\dfrac{7}{6}\times\dfrac{8}{7}\times\cdot\cdot\cdot\times\dfrac{2020}{2019}\times\dfrac{2021}{2020}\)
\(F=\dfrac{6\times7\times8\times\cdot\cdot\cdot\times2020\times2021}{5\times6\times7\times\cdot\cdot\cdot\times2019\times2020}\)
\(F=\dfrac{2021}{5}\)
| \(Huyền\) |
C
15 tháng 5 2022
\(f=1^1_5\times1^1_6\times1^1_7\times......\times1^1_{2019}\times1^1_{2022}\)
\(f=\dfrac{6}{5}\times\dfrac{7}{6}\times\dfrac{8}{7}\times....\times\dfrac{2020}{2019}\times\dfrac{2021}{2020}\)
\(f=\dfrac{6\times7\times8\times....\times2020\times2021}{5\times6\times7\times.....\times2019\times2020}\)
\(f=\dfrac{2021}{5}\)
\(#Tarus\)
HN
1
28 tháng 4 2020
Ta có: 1 + ( 1 + 2 ) + ( 1 + 2 + 3 ) + ... + ( 1 + 2 + 3 +...+ 2020)
= ( 1 + 1 + 1 +... + 1 ) + (2 + 2 +...+ 2 ) + ( 3 + 3+...+ 3 ) + ...+ 2020
Có 2020 số 1 ; 2019 số 2 ; 2018 số 3 ;... ; 1 số 2020
= 2020 x 1 + 2019 x 2 + 2018 x 3 + ... + 2020x 1
=> \(M=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+2020\right)}{1\times2020+2\times2019+...+2020\times1}\)
= \(\frac{1\times2020+2\times2019+...+2020\times1}{1\times2020+2\times2019+...+2020\times1}=1\)
CT
25 tháng 5 2021
( 1 - 1 / 2 ) x ( 1 - 1 / 3 ) x K x (1 - 1 / 2019 ) x (1 - 1 / 2020 )
Chỗ chứ K đấy tớ thay bằng ... nhé
( 1 - \(\frac{1}{2}\) ) x ( 1 - \(\frac{1}{3}\) ) x ... x ( 1 - \(\frac{1}{2019}\) ) x ( 1 - \(\frac{1}{2020}\) )
= ( \(\frac{2}{2}\) - \(\frac{1}{2}\) ) x ( \(\frac{3}{3}\) - \(\frac{1}{3}\) ) x...x ( \(\frac{2019}{2019}\) - \(\frac{1}{2019}\) ) x ( \(\frac{2020}{2020}\) - \(\frac{1}{2020}\) )
= \(\frac{1}{2}\) x \(\frac{2}{3}\) x...x \(\frac{2018}{2019}\) x \(\frac{2019}{2020}\)
= \(\frac{1.2....2018.1019}{2.3.....2019.2020}\)
= \(\frac{1}{2020}\)
Học tốt !
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)..\left(1+\frac{1}{2019}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{2020}{2019}\)
\(=\frac{2020}{2}=1010\)
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{2019}\right)\)
\(=\frac{3}{2}\cdot\frac{4}{3}\cdot...\cdot\frac{2020}{2019}\)
\(=\frac{2020}{2}\)
\(=1010\)