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A=\(1\frac{1}{2}.1\frac{1}{3}.1\frac{1}{4}....1\frac{1}{2015}\)
A=\(\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{2016}{2015}\)
A=\(\frac{3.4.5.....2016}{2.3.4....2015}\)
A=\(\frac{2016}{2}=1008\)
nhanh tay lên các bạn ai trả lời đầu tiên mình cho 5 cái tick
\(A=1\frac{1}{2}.1\frac{1}{3}....1\frac{1}{2015}\)
\(=\frac{3}{2}.\frac{4}{3}......\frac{2016}{2015}\)
\(=\frac{2016}{2}=1008\)
\(=\frac{3}{2}\times\frac{4}{3}\times.............\times\frac{2016}{2015}\)
\(=\frac{3\times4\times.............\times2016}{2\times3\times..............\times2015}\)
\(=\frac{2016}{2}=1008\)
Ta có : \(1\frac{1}{2}\times1\frac{1}{3}\times1\frac{1}{4}\times.....\times1\frac{1}{2017}\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{2018}{2017}\)
\(=\frac{2018}{2}=1009\)
\(1\frac{1}{2}\times1\frac{1}{3}\times1\frac{1}{4}\times....\times1\frac{1}{2017}\)
\(=\frac{3}{2}\times\frac{4}{3}\times\frac{5}{4}\times...\times\frac{2018}{2017}\)
\(=\frac{3\times4\times5\times.....\times2018}{2\times3\times4\times.....\times2017}\)
\(=\frac{2018}{2}=1009\)
1, =\(\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}=\frac{1}{2}\)
2, A=\(\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{99}{100}\)
= \(\frac{1\cdot2\cdot3\cdot....\cdot99}{2\cdot3\cdot4\cdot...\cdot100}=\frac{1}{100}\)
Vậy ......
hok 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 ~
có tử bằng 1+(1+2)+(1+2+3)+...+(1+2+3+...+98)
vậy sẽ có 98 lần số 1 97 lần số 2 96 lần số 3 ... và 1 lần số 98
=> Tử bằng 1x98 + 2x97 + ... + 98x1 = mẫu
=> B=1
\(A=1\frac{1}{2}\cdot1\frac{1}{3}\cdot1\frac{1}{4}\cdot...\cdot1\frac{1}{2015}\)
\(\Rightarrow A=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot...\cdot\frac{2016}{2015}\)
\(\Rightarrow A=\frac{3\cdot4\cdot5\cdot...\cdot2016}{2\cdot3\cdot4\cdot...\cdot2015}=\frac{2016}{2}=1008\)
Vậy A = 1008
a=1008