tính:
a,\(A=\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot\left(\frac{1}{4}+1\right)\cdot...\cdot\left(\frac{1}{99}+1\right)\)
Những câu hỏi liên quan
tính nhanh a, frac{-2}{5}cdotleft(frac{5}{17}-frac{9}{15}right)-frac{2}{5}cdotfrac{2}{17}+frac{-2}{5}b, frac{1}{5}cdotleft(frac{4}{13}-frac{9}{11}right)+frac{1}{3}left(frac{9}{13}-frac{4}{22}right)c, left(frac{1}{2}+1right)cdotleft(frac{1}{3}+1right)cdotleft(frac{1}{4}+1right)cdot...cdotleft(frac{1}{99}+1right)d, left(1-frac{1}{2}right)cdotleft(1-frac{1}{3}right)cdotleft(1-frac{1}{4}right)cdot...cdotleft(1-frac{1}{100}right)
Đọc tiếp
tính nhanh
a, \(\frac{-2}{5}\cdot\left(\frac{5}{17}-\frac{9}{15}\right)-\frac{2}{5}\cdot\frac{2}{17}+\frac{-2}{5}\)
b, \(\frac{1}{5}\cdot\left(\frac{4}{13}-\frac{9}{11}\right)+\frac{1}{3}\left(\frac{9}{13}-\frac{4}{22}\right)\)
c, \(\left(\frac{1}{2}+1\right)\cdot\left(\frac{1}{3}+1\right)\cdot\left(\frac{1}{4}+1\right)\cdot...\cdot\left(\frac{1}{99}+1\right)\)
d, \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{100}\right)\)
Mk ko biết lm nhưng cứ k thoải mái nha
SORRY
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\(\left(1+\frac{1}{2}\right)\cdot\left(1+\frac{1}{3}\right)\cdot\left(1+\frac{1}{4}\right)\cdot....\cdot\left(1+\frac{1}{98}\right)\cdot\left(1+\frac{1}{99}\right)=\)?
\(\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)...\left(1+\frac{1}{99}\right)=\frac{3}{2}.\frac{4}{3}...\frac{100}{99}=\frac{100}{2}=50\)
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= \(\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot\cdot\cdot\cdot\frac{99}{98}\cdot\frac{100}{99}=\frac{3.4.5....99.100}{2.3.4....98.99}=\frac{100}{2}=50\)
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\(\left(1+\frac{1}{2}\right).\left(1+\frac{1}{3}\right).\left(1+\frac{1}{4}\right).........\left(1+\frac{1}{99}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{99}{98}.\frac{100}{99}\)
\(=\frac{3.4.5....99.100}{2.3.4.5....98.99}\)
\(=\frac{100}{2}\)
\(=50\)
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Tinh \(B=\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\left(\frac{1}{4^2}-1\right)\cdot....\cdot\left(\frac{1}{98^2}-1\right)\cdot\left(\frac{1}{99^2}-1\right)\)
Tính giá trị biểu thức :
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{99}\right)\cdot\left(1-\frac{1}{100}\right)\)
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
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Tính
A=\(\left(1-\frac{1}{21}\right)\cdot\left(1-\frac{1}{28}\right)\cdot\left(1-\frac{1}{36}\right)\cdot....\cdot\left(1-\frac{1}{1326}\right)\)
B=\(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\left(1+\frac{1}{3\cdot5}\right)\cdot....\cdot\left(1+\frac{1}{99\cdot101}\right)\)
Tính các tích sau: với n là số tự nhiên, n<3
a) \(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{n}\right)\)
b) \(\left(1-\frac{1}{2^2}\right)\cdot\left(1-\frac{1}{3^2}\right)\cdot\left(1-\frac{1}{4^2}\right)\cdot...\cdot\left(1-\frac{1}{n^2}\right)\)
\(C=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot....\cdot\left(1-\frac{2}{99\cdot100}\right)\)
\(\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot\left(1-\frac{1}{5}\right)\cdot...\cdot\left(1-\frac{1}{2017}\right)\cdot\left(1-\frac{1}{2018}\right)\)
= (1/2).(2/3).(4/5).(5/6)......(2016/2017).(2017/2018)
=1.2.3.4.5......2016.2017/2.3.4.5.....2017.2018
=1/2018
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\(\left(1-\frac{1}{2}\right)\left(1-\frac{1}{3}\right)\left(1-\frac{1}{4}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\frac{1}{2017}\right)\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\cdot\cdot\frac{2016}{2017}\cdot\frac{2017}{2018}\)
\(=\frac{1\cdot2\cdot3\cdot\cdot\cdot\cdot\cdot2016\cdot2017}{2\cdot3\cdot4\cdot\cdot\cdot\cdot2017\cdot2018}\)
\(=\frac{1}{2018}\)
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\(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)....\left(1-\frac{1}{2017}\right)\left(1-\frac{1}{2018}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2016}{2017}.\frac{2017}{2018}\)
\(=\frac{1}{2018}\)
p/s: chúc bạn hok tốt
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Xem thêm câu trả lời
TÍNH
\(C=\left(1+\frac{2}{3}\right)\cdot\left(1+\frac{2}{5}\right)\cdot\left(1+\frac{2}{7}\right)\cdot\cdot\cdot\cdot\cdot\left(1+\frac{2}{2015}\right)\cdot\left(1+\frac{2}{2017}\right)\)
\(D=\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{6}\right)\cdot\left(1-\frac{1}{10}\right)\cdot\left(1-\frac{1}{15}\right)\cdot\cdot\cdot\cdot\left(1-\frac{1}{780}\right)\)
\(C=\frac{5}{2}\cdot\frac{7}{5}\cdot\frac{9}{7}\cdot\frac{11}{9}\cdot...\cdot\frac{2017}{2015}\cdot\frac{2019}{2017}=\frac{2019}{2}\)
\(D=\left(1-\frac{1}{\frac{2\cdot3}{2}}\right)\cdot\left(1-\frac{1}{\frac{3\cdot4}{2}}\right)\cdot\left(1-\frac{1}{\frac{4\cdot5}{2}}\right)\cdot\left(1-\frac{1}{\frac{5\cdot6}{2}}\right)\cdot...\cdot\left(1-\frac{1}{\frac{39\cdot40}{2}}\right)\)
\(=\left(1-\frac{2}{2\cdot3}\right)\cdot\left(1-\frac{2}{3\cdot4}\right)\cdot\left(1-\frac{2}{4\cdot5}\right)\cdot\left(1-\frac{2}{5\cdot6}\right)\cdot...\cdot\left(1-\frac{2}{39\cdot40}\right)\cdot\)
Nhận xét: \(1-\frac{2}{n\left(n+1\right)}=\frac{n\left(n+1\right)-2}{n\left(n+1\right)}=\frac{n^2+n-2}{n\left(n+1\right)}=\frac{\left(n+2\right)\left(n-1\right)}{n\left(n+1\right)}\)nên:
\(D=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\frac{6\cdot3}{4\cdot5}\cdot\frac{7\cdot4}{5\cdot6}\cdot\frac{8\cdot5}{6\cdot7}\cdot...\cdot\frac{41\cdot38}{39\cdot40}=\)
\(D=\frac{4\cdot5\cdot6\cdot7\cdot...\cdot41\times1\cdot2\cdot3\cdot4\cdot...\cdot38}{2\cdot3\cdot4\cdot5\cdot...\cdot39\times3\cdot4\cdot5\cdot6\cdot..\cdot40}=\frac{1}{39}\cdot\frac{41}{3}=\frac{41}{117}\)
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