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TT
1
1 tháng 6 2019
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2019.2020}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2019}-\frac{1}{2020}\)
\(A=1-\frac{1}{2020}\)
\(A=\frac{2019}{2020}\)
1 tháng 6 2019
\(B=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{2017.2019}\)
\(2B=\frac{2}{1.3}+\frac{2}{3.5}=\frac{2}{5.7}+...+\frac{2}{2017.2019}\)
\(2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}=\frac{1}{5}-\frac{1}{7}+...+\frac{1}{2017}-\frac{1}{2019}\)
\(2B=1-\frac{1}{2019}\)
\(2B=\frac{2018}{2019}\)
\(B=\frac{2018}{2019}:2=\frac{1009}{2019}\)
VN
20 tháng 1 2023
a)
\(11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(=\dfrac{146}{13}-\left(\dfrac{18}{7}+\dfrac{68}{13}\right)\)
\(=\dfrac{146}{13}-\dfrac{18}{7}-\dfrac{68}{13}\)
\(=\left(\dfrac{146}{13}-\dfrac{68}{13}\right)-\dfrac{18}{7}\)
\(=6-\dfrac{18}{7}\)
\(=\dfrac{24}{7}\)
b)
\(\dfrac{2}{7}\times5\dfrac{1}{4}-\dfrac{2}{7}\times3\dfrac{1}{4}\)
\(=\dfrac{2}{7}\times\dfrac{21}{4}-\dfrac{2}{7}\times\dfrac{13}{4}\)
\(=\dfrac{2}{7}\times\left(\dfrac{21}{4}-\dfrac{13}{4}\right)\)
\(=\dfrac{2}{7}\times2\)
\(=\dfrac{4}{7}\)
Y
20 tháng 1 2023
\(a,11\dfrac{3}{13}-\left(2\dfrac{4}{7}+5\dfrac{3}{13}\right)\)
\(=\dfrac{146}{13}-\left(\dfrac{18}{7}+\dfrac{68}{13}\right)\)
\(=\dfrac{146}{13}-\dfrac{68}{13}-\dfrac{18}{7}\)
\(=6-\dfrac{18}{7}\)
\(=\dfrac{24}{7}\)
\(b,\dfrac{2}{7}\times5\dfrac{1}{4}-\dfrac{2}{7}\times3\dfrac{1}{4}\)
\(=\dfrac{2}{7}\times\dfrac{21}{4}-\dfrac{2}{7}\times\dfrac{13}{4}\)
\(=\dfrac{2}{7}\times\left(\dfrac{21}{4}-\dfrac{13}{4}\right)\)
\(=\dfrac{2}{7}\times2\)
\(=\dfrac{4}{7}\)
D
28 tháng 4 2017
Bài 1:
\(\frac{1}{2}.\frac{1}{3}+\frac{1}{3}.\frac{1}{4}+....+\frac{1}{8}.\frac{1}{9}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{8.9}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+.....+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{2}-\frac{1}{9}=\frac{7}{18}\)
28 tháng 4 2017
\(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{x\left(x+1\right):2}=1\frac{2016}{2017}\)
\(\Rightarrow\frac{1}{2}\left(1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{4033}{2017}\)
\(\Leftrightarrow\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}=\frac{4033}{4034}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{4033}{4034}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{4033}{4034}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{4033}{4034}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{4033}{4034}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{4034}\)
\(\Rightarrow x+1=4034\)
\(\Rightarrow x=4034-1\)
\(\Rightarrow x=4033\)
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