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23 tháng 5 2018
\(\frac{242}{363}+\frac{1616}{2121}=\frac{2}{7}.\frac{2015}{A}\)
\(\frac{11.11.2}{11.11.3}+\frac{101.16}{101.21}=\frac{2}{7}.\frac{2015}{A}\)
\(\frac{2}{3}+\frac{16}{21}=\frac{2}{7}.\frac{2015}{A}\)
\(\Rightarrow\frac{2}{7}.\frac{2015}{A}=\frac{10}{7}\)
\(\frac{2015}{A}=\frac{10}{7}:\frac{2}{7}\)
\(\frac{2015}{A}=5\)
\(A=2015:5\)
\(A=403\)
23 tháng 5 2018
\(\left(\frac{242}{363}+\frac{1616}{2121}\right)=\frac{2}{7}.\frac{2015}{A}\)
\(\Leftrightarrow\left(\frac{2}{3}+\frac{16}{21}\right)=\frac{2}{7}.\frac{2015}{A}\)
\(\Leftrightarrow\frac{10}{7}=\frac{2}{7}.\frac{2015}{A}\)
\(\Leftrightarrow5=\frac{2015}{A}\)
\(\Leftrightarrow A=403\)
D
21 tháng 7 2018
\(\left(2013.2014+2014.2015+2015.2016\right)\left(1+\frac{1}{3}-1\frac{1}{3}\right)\)
\(=\left(2013.2014+2014.2015+2015.2016\right)\left(\frac{4}{3}-\frac{4}{3}\right)\)
\(=\left(2013.2014+2014.2015+2015.2016\right).0\)
\(=0\)
NN
18 tháng 8 2017
= \(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{19}{20}=\frac{1}{20}\)
18 tháng 8 2017
=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}......\frac{19}{20}\)
= đến đây bn xem trong thống kê hỏi đáp nhé
= :))
7 tháng 8 2017
a. ta có (0.1+0.19)+(0.2+0.18)......+0.10
A=0.20+0.20++0.20+0.20+0.20+0.20+0.20+0.20+0.20+0.10
A=1.90
câu b mình pó tay
TH
7 tháng 8 2017
a ) \(A=0,1+0,2+...+0,19\)
\(A=\left(0,1+0,2+...+0,9\right)+\left(0,10+0,11+...+0,19\right)\)
\(A=0,1\times\left(1+2+...+9\right)+0,1\times\left(1+1,1+...+1,9\right)\)
\(A=0,1\times45+0,1\times14,5\)
\(A=0,1\times\left(45+14,5\right)\)
\(A=0,1\times59,5\)
\(A=5,95\)
b ) \(B=\left(2017\times2016+2014\times2015\right)\times\left(1+\frac{1}{2}\div1\frac{1}{2}+1\frac{1}{3}\right)\)
\(B=\left(2017\times2016+2014\times2015\right)\times\left(1+\frac{1}{2}\div\frac{3}{2}+\frac{4}{3}\right)\)
\(B=\left(2017\times2016+2014\times2015\right)\times\left(1+\frac{2}{6}+\frac{4}{3}\right)\)
\(B=\left(2017\times2016+2014\times2015\right)\times\frac{8}{3}\)
23 tháng 1 2016
(1-1/2).(1-1/3).(1-1/4).(1-1/5)=b/100
=> 1/2.2/3.3/4.4/5=b/100
=> 1/5=b/100
=> b=100:5=20
DD
6 tháng 8 2015
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{99}\right)=\frac{3}{2}.\frac{4}{3}...\frac{100}{99}=\frac{100}{2}=50\)
6 tháng 8 2015
= \(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\cdot\cdot\cdot\frac{99}{98}\cdot\frac{100}{99}=\frac{3.4.5....99.100}{2.3.4....98.99}=\frac{100}{2}=50\)
T
20 tháng 6 2019
#)Giải :
\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)
\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\left(\frac{1}{2}+\frac{1}{2}\right)\)
\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2016}+\frac{2009}{2018}\right)\times0\)
\(=0\)
20 tháng 6 2019
\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{1}{3}+\frac{1}{2}\right)\)
\(=\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).\left(\frac{1}{6}+\frac{2}{6}+\frac{3}{6}\right)\)
=\(\left(\frac{2012}{2015}+\frac{2011}{2016}+\frac{2010}{2017}+\frac{2009}{2018}\right).0\)
\(=0\)
HP
1 tháng 8 2016
Ta có:
\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{99}\right).\left(1-\frac{1}{100}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}...\frac{98}{99}.\frac{99}{100}\) \(=\frac{1.2.3...98.99}{2.3.4...99.100}=\frac{1}{100}\)
nha
HT
3 tháng 7 2018
Câu b:
\(\frac{21}{8}:\frac{5}{6}+\frac{1}{2}:\frac{5}{6}\)
= \(\frac{63}{20}+\frac{3}{5}\)
= \(\frac{15}{4}\)
LT
7 tháng 7 2018
\(\left(\frac{21}{8}+\frac{1}{2}\right):\frac{5}{6}\)
\(\frac{25}{8}:\frac{5}{6}\)
\(\frac{25}{8}.\frac{6}{5}\)
\(\frac{30}{8}\)
\(\left(1,256.a+1,743.a\right)-2014=2015\)
\(\Leftrightarrow a\left(1,256+1,743\right)=2015+2014\)
\(\Leftrightarrow a.2,999=4029\)
\(\Leftrightarrow a=\frac{4029}{2,999}=1343,\left(44781593\right)\)
\(\left(1,256.a+1,743.a\right)-2014=2015\)
\(=\left(1,256.a+1,743.a\right)=2015+2014=4029\)
\(=\left(1,256.a+1,743.a\right)=4029\)
\(=\)gộp lại rùi tính