\(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)\left(1-\dfrac{1}{15}\right)...\left(1-\dfrac{1}{780}\right)\)
Những câu hỏi liên quan
Tính:
\(A=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)
\(A=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)......\left(1-\dfrac{1}{780}\right)\)
= \(\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.......\dfrac{779}{780}\)
= \(\dfrac{4}{6}.\dfrac{10}{12}.\dfrac{18}{20}.....\dfrac{1558}{1560}\)
= \(\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}......\dfrac{38.41}{39.40}\)
= \(\dfrac{1.4.2.5.3.6.....38.41}{2.3.3.4.4.5...39.40}\)
= \(\dfrac{\left(1.2.3....38\right)\left(4.5.6....41\right)}{\left(2.3.4....39\right)\left(3.4.5...40\right)}\)
= \(\dfrac{1}{39}.\dfrac{41}{3}\) = \(\dfrac{41}{117}\)
Đúng 0
Bình luận (1)
tính
\(B=\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{6}\right).\left(1-\dfrac{1}{10}\right).........\left(1-\dfrac{1}{780}\right)\)
\(B=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)
\(B=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}...\dfrac{779}{780}=\dfrac{4}{6}.\dfrac{10}{12}.\dfrac{18}{20}...\dfrac{1558}{1560}\)
\(B=\dfrac{1.4}{2.3}.\dfrac{2.5}{3.4}.\dfrac{3.6}{4.5}.....\dfrac{38.41}{39.40}\)
\(B=\dfrac{\left(1.2.3.....38\right)\left(4.5.6.....41\right)}{\left(2.3.4.....39\right)\left(3.4.5.....40\right)}=\dfrac{1.41}{39.3}=\dfrac{41}{117}\)
Đúng 0
Bình luận (0)
\(\dfrac{\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}\right):\left(\dfrac{1}{6}+\dfrac{1}{10}-\dfrac{1}{15}\right)}{\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}-\dfrac{1}{5}\right):\left(\dfrac{1}{4}-\dfrac{1}{6}\right)}=\)
\(\dfrac{\left(\dfrac{5}{30}+\dfrac{3}{30}+\dfrac{2}{30}\right):\left(\dfrac{5}{30}+\dfrac{3}{30}-\dfrac{2}{30}\right)}{\left(\dfrac{30}{60}-\dfrac{20}{60}+\dfrac{15}{60}-\dfrac{12}{60}\right):\left(\dfrac{3}{12}-\dfrac{2}{12}\right)}=\dfrac{\dfrac{1}{3}:\dfrac{1}{5}}{\dfrac{13}{60}:\dfrac{1}{12}}=\dfrac{\dfrac{1}{3}\times5}{\dfrac{13}{60}\times12}=\dfrac{\dfrac{5}{3}}{\dfrac{13}{5}}=\dfrac{25}{39}\)
Đúng 1
Bình luận (1)
=\(\dfrac{\dfrac{1}{3}:\dfrac{1}{5}}{\dfrac{13}{60}:\dfrac{1}{12}}=\dfrac{\dfrac{5}{3}}{\dfrac{13}{5}}=\dfrac{25}{39}\)
Đúng 1
Bình luận (1)
Tính:
B=\(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)\left(1-\dfrac{1}{15}\right)...\left(1-\dfrac{1}{2450}\right)\)
Tính B = \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)\left(1-\dfrac{1}{15}\right)...\left(1-\dfrac{1}{2450}\right)\)
1. 25dfrac{1}{7}:left(-dfrac{5}{7}right)-15dfrac{1}{7}:left(-dfrac{5}{7}right)+dfrac{4}{5} 3. 2dfrac{2}{3}:left{left[left(3,72-0.02right)dfrac{10}{37}right]:dfrac{5}{6}+2,8right}-dfrac{7}{15}2. left(3+dfrac{4}{5}-dfrac{5}{12}right)left(dfrac{6}{7}-dfrac{3}{5}right)^24.23+3.left(-dfrac{1}{2}right)^2-22.4+left[left(-2right)^2:dfrac{1}{2}right]
Đọc tiếp
1. \(25\dfrac{1}{7}:\left(-\dfrac{5}{7}\right)-15\dfrac{1}{7}:\left(-\dfrac{5}{7}\right)+\dfrac{4}{5}\) 3. \(2\dfrac{2}{3}:\left\{\left[\left(3,72-0.02\right)\dfrac{10}{37}\right]:\dfrac{5}{6}+2,8\right\}-\dfrac{7}{15}\)
2. \(\left(3+\dfrac{4}{5}-\dfrac{5}{12}\right)\left(\dfrac{6}{7}-\dfrac{3}{5}\right)^2\)
4.23+3.\(\left(-\dfrac{1}{2}\right)^2\)-22.4+\(\left[\left(-2\right)^2:\dfrac{1}{2}\right]\)
2: \(=\dfrac{203}{60}\cdot\dfrac{81}{1225}=\dfrac{783}{3500}\)
Đúng 0
Bình luận (0)
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{5}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{7}\right)\left(1-\dfrac{1}{8}\right)\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{10}\right)\)
\(\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right).......\left(1-\dfrac{1}{10}\right)\)
\(=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right).........\left(\dfrac{10}{10}-\dfrac{1}{10}\right)\)
\(=\dfrac{1}{2}.\dfrac{2}{3}......\dfrac{9}{10}\)
\(=\dfrac{1}{10}\)
Đúng 0
Bình luận (0)
25 tháng 3 2017 lúc 15:44
C=\(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)\left(1-\dfrac{1}{15}\right).....\left(1-\dfrac{1}{210}\right)\)
Tính C
\(C=\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)...\left(1-\dfrac{1}{210}\right)\)
\(=\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot\cdot\cdot\dfrac{209}{210}\)
Đúng 0
Bình luận (0)
Tính
Aleft(dfrac{1}{3}-1right)left(dfrac{1}{6}-1right)left(dfrac{1}{10}-1right)left(dfrac{1}{15}-1right)left(dfrac{1}{21}-1right)left(dfrac{1}{28}-1right)left(dfrac{1}{36}-1right)
Bleft(1-dfrac{1}{2^2}right)left(1-dfrac{1}{3^2}right)........left(1-dfrac{1}{10^2}right)
C1+dfrac{1}{2}+dfrac{1}{2^2}+dfrac{1}{2^3}+..........+dfrac{1}{2^{2016}}
Giúp mk nha!Cảm ơn rất nhìu!
Đọc tiếp
Tính
A=\(\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{6}-1\right)\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{15}-1\right)\left(\dfrac{1}{21}-1\right)\left(\dfrac{1}{28}-1\right)\left(\dfrac{1}{36}-1\right)\)
B=\(\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)........\left(1-\dfrac{1}{10^2}\right)\)
C=\(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+..........+\dfrac{1}{2^{2016}}\)
Giúp mk nha!Cảm ơn rất nhìu!
Ta có: \(A=\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{6}-1\right)\left(\dfrac{1}{10}-1\right)\left(\dfrac{1}{15}-1\right)\left(\dfrac{1}{21}-1\right)\left(\dfrac{1}{28}-1\right)\left(\dfrac{1}{36}-1\right)\)
\(=\dfrac{-2}{3}.\dfrac{-5}{6}.\dfrac{-9}{10}.\dfrac{-14}{15}.\dfrac{-20}{21}.\dfrac{-27}{28}.\dfrac{-35}{36}\)
\(=\dfrac{-2.\left(-5\right).3.\left(-3\right).2.\left(-7\right).\left(-4\right).5.\left(-3\right).9.5.\left(-7\right)}{3.2.3.2.5.3.5.3.7.4.7.4.9}\)
\(=\dfrac{-5}{3.4}=\dfrac{-5}{12}\)
Vậy \(A=\dfrac{-5}{12}.\)
Đúng 0
Bình luận (0)
\(C=1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\)
\(2C=2\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\right)\)
\(2C=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+....+\dfrac{1}{2^{2015}}\)
\(2C-C=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2015}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2016}}\right)\)
\(C=2-\dfrac{1}{2^{2016}}\)
Đúng 0
Bình luận (5)