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\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2012}}\)
\(2A=2+1+\frac{1}{2}+...+\frac{1}{2^{2011}}\)
\(2A-A=(2+1+\frac{1}{2}+...+\frac{1}{2^{2011}})-(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2012}})\)
\(A=2-\frac{1}{2^{2012}}\)
Vậy A = \(2-\frac{1}{2^{2012}}\)
~Chúc bạn học tốt~
Xét\(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\)
Lấy 2A - A Ta được
\(A=2-\frac{1}{2^{2012}}\)
\(1\frac{1}{2}.1\frac{1}{3}.1\frac{1}{4}.....1\frac{1}{2015}\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}........\frac{2016}{2015}\)
\(=\frac{3.4.5.....2016}{2.3.4....2015}=\frac{2016}{2}=1008\)
\(A=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}...\frac{2016}{2015}\)
\(A=\frac{2016}{2}=1008\)
Xong nhé bạn!
Tờ làm luôn, ko ghi đề nữa nhé
\(A=\frac{\frac{24}{12}-\frac{4}{12}+\frac{3}{12}}{\frac{24}{12}+\frac{2}{12}-\frac{3}{12}}\)
\(A=\frac{\frac{23}{12}}{\frac{23}{12}}=1\)
Vậy A=1
\(A=\frac{2-\frac{1}{3}+\frac{1}{4}}{2+\frac{1}{6}-\frac{1}{4}}\)\(=\frac{2-\frac{2}{6}+\frac{2}{8}}{2+\frac{2}{12}-\frac{2}{8}}\)\(=\frac{2\left(1-\frac{1}{6}+\frac{1}{8}\right)}{-2\left(1-\frac{1}{12}+\frac{1}{8}\right)}\)\(=-1\)
c)\(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2012}}\)
\(2A=2\left(1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{2012}}\right)\)
\(2A=2+1+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2011}}\)
\(2A-A=\left(2+1+\frac{1}{2^2}+\frac{1}{2^3}+....+\frac{1}{2^{2011}}\right)-\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+....\frac{1}{2^{2012}}\right)\)
\(A=2-\frac{1}{2^{2012}}\)
1/
A=1/1-1/2+1/2-1/3+1/3-1/4+...+1/99-1/100
A=1/1-1/100
Vì 1/100>0
-->1/1-1/100<1
-->A<1
\(T=\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right).......\left(\frac{1}{98}+1\right).\left(\frac{1}{99}+1\right)\)
\(T=\left(\frac{1}{2}+\frac{2}{2}\right).\left(\frac{1}{3}+\frac{3}{3}\right).\left(\frac{1}{4}+\frac{4}{4}\right).....\left(\frac{1}{98}+\frac{98}{98}\right).\left(\frac{1}{99}+\frac{99}{99}\right)\)
\(T=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{99}{98}.\frac{100}{99}\)
\(T=\frac{3.4.5....99.100}{2.3.4.....98.99}\)
\(T=\frac{100}{2}\)
\(T=50\)
Vậy T = 50
Chúc bạn học tốt!
\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{1+\left(1+\frac{2016}{2}\right)+\left(1+\frac{2015}{3}\right)+...+\left(1+\frac{1}{2017}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2018}}\)
\(A=2018\)
Ta có :
\(A=\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{\left(\frac{2017}{1}-1-1-...-1\right)+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=\frac{2018\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(A=2018\)
Vậy \(A=2018\)
Chúc bạn học tốt ~
=1\99-(1\1.2+1\1.3+....1\98.99)
=1\99-(1-1\2+1\2-1\3+1\3...+1\98-1\99)
=1\99-(1-1\99)
=1\99-1\98
=-97\99
\(D=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2014}{2015}\)
\(D=\frac{1\cdot2\cdot3\cdot...\cdot2014}{2\cdot3\cdot4\cdot...\cdot2015}=\frac{1}{2015}nhebn\)
(2/2-1/2).(3/3-1/3).(4/4-1/4)............(2015/2015-1/2015 )
1/2.2/3.3/4.....................2014/2015
=1/2015