\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right).....\left(1-\frac{1}{2007^2}\right)\)
mong mấy bạn giúp.
Những câu hỏi liên quan
Thực hiện phép tính :a, sqrt{12frac{1}{4}}.left(-frac{2}{7}right)^2-left[2,left(4right).2frac{5}{11}right]:(-frac{42}{5})b, left(1-frac{1}{2}right).left(1-frac{1}{3}right).left(1-frac{1}{4}right)...left(1-frac{1}{2007}right)c, frac{0,375-0,3+frac{3}{11}+frac{3}{12}}{-0,625+0,5-frac{5}{11}-frac{5}{12}}+frac{1,5+1-0,75}{2,5+frac{5}{3}-1,25}Các bạn giải nhanh giúp mình nha!
Đọc tiếp
Thực hiện phép tính :
a, \(\sqrt{12\frac{1}{4}}.\left(-\frac{2}{7}\right)^2-\left[2,\left(4\right).2\frac{5}{11}\right]:(-\frac{42}{5})\)
b, \(\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2007}\right)\)
c, \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{-0,625+0,5-\frac{5}{11}-\frac{5}{12}}+\frac{1,5+1-0,75}{2,5+\frac{5}{3}-1,25}\)
Các bạn giải nhanh giúp mình nha!
Tính
\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2007^2}\right)\)
\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2007^2}\right)\)
\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}......\frac{2007^2-1}{2007^2}\)
\(=\frac{\left(2-1\right)\left(2+1\right)}{2.2}.\frac{\left(3-1\right)\left(3+1\right)}{3.3}.\frac{\left(4-1\right)\left(4+1\right)}{4.4}....\frac{\left(2007-1\right)\left(2007+1\right)}{2007.2007}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{2006.2008}{2007.2007}\)
\(=\frac{\left(1.2.3.....2006\right)\left(3.4.5....2008\right)}{\left(2.3.4...2007\right)\left(2.3.4.....2007\right)}\)
\(=\frac{1.2008}{2007.2}=\frac{1004}{2007}\)
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Tính B = \(1+\frac{1}{2}.\left(1+2\right)+\frac{1}{3}.\left(1+2+3\right)+\frac{1}{4}.\left(1+2+3+4\right)+....+\frac{1}{20}.\left(1+2+3+...+20\right)\)
Phong cần gấp
Mong đc giúp đỡ
áp dung \(1+2+...+n=\frac{n\left(n+1\right)}{2}\)
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left(frac{-5}{12}+frac{7}{4}-frac{3}{8}right)-left[4frac{1}{2}-7frac{1}{3}right]-left(frac{1}{4}-frac{5}{2}right)left[2frac{1}{4}-5frac{3}{2}right]-left(frac{3}{10}-1right)-5frac{1}{2}+left(frac{1}{3}-frac{5}{6}right)frac{4}{7}-left(3frac{2}{5}-1frac{1}{2}right)-frac{5}{21}+left[3frac{1}{2}-4frac{2}{3}right]frac{1}{8}-1frac{3}{4}+left(frac{7}{8}-3frac{7}{2}+frac{3}{4}right)-left[frac{7}{4}-frac{5}{8}right]left(frac{3}{5}-2frac{1}{10}+frac{11}{20}right)-left[frac{-3}{4}+1frac{7}{2}right]left[-2fr...
Đọc tiếp
\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)
\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)
\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)
\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)
\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)
\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)
\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)
\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)
\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)
\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)
\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)
\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)
\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)
\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)
\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)
\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)
\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)
\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)
TRÌNH BÀY GIÚP MÌNH NHA
Tính \(S=\left(1+\frac{1}{2}\right)\left(1+\frac{1}{2}+\frac{1}{3}\right)\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+...+\left(1+\frac{1}{2}+...+\frac{1}{10}\right)\)
Các bạn giúp mình nhé !~~~
Tính (và nêu cách tính):
\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)........\left(1-\frac{1}{2007^2}\right)\)
mấy bạn giải giúp mk với B=\(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{20}\left(1+..+20\right)\)
\(Giải:\)
\(B=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+.......+\frac{1}{20}\left(1+.......+20\right)\)
\(B=1+\frac{1}{2}\left(\frac{3.2}{2}\right)+\frac{1}{3}\left(\frac{4.3}{2}\right)+\frac{1}{4}\left(\frac{5.4}{2}\right)+.......+\frac{1}{20}\left(\frac{21.20}{2}\right)\)
\(B=1+\frac{3.2}{2.2}+\frac{4.3}{3.2}+\frac{5.4}{4.2}+.........+\frac{21.20}{20.2}\)
\(B=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+.........+\frac{21}{2}=\frac{1}{2}+\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+.........+\frac{21}{2}-\frac{1}{2}\)
\(=\frac{1+2+3+.........+21-1}{2}=\frac{22.21-1}{4}=\frac{242-1}{4}=\frac{241}{4}=60\frac{1}{4}\)
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a)\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)x......x\left(1-\frac{1}{18}\right)x\left(1-\frac{1}{19}\right)x\left(1-\frac{1}{20}\right)\)
b)\(1\frac{1}{2}x1\frac{1}{3}x1\frac{1}{4}x1\frac{1}{5}x......x1\frac{1}{2005}x1\frac{1}{2006}x1\frac{1}{2007}\)
\(x\)là dấu nhân hả bạn? Nếu vậy thì mk làm cho nhé
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.......\cdot\frac{17}{18}\cdot\frac{18}{19}\cdot\frac{19}{20}=\frac{1}{20}\)
Vậy \(A=\frac{1}{20}\)
\(B=1\frac{1}{2}\cdot1\frac{1}{3}\cdot1\frac{1}{4}\cdot........\cdot1\frac{1}{2005}\cdot1\frac{1}{2006}\cdot1\frac{1}{2007}\)
\(B=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot......\cdot\frac{2006}{2005}\cdot\frac{2007}{2006}\cdot\frac{2008}{2007}=\frac{2008}{2}=1004\)
Vậy \(B=1004\)
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DẤU CHẤM LÀ DẤU NHÂN
a,
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{19}{20}=\frac{1}{20}\)
b, \(1\frac{1}{2}.1\frac{1}{3}....1\frac{1}{2017}=\frac{3}{2}.\frac{4}{3}....\frac{2018}{2017}=\frac{2018}{2}=1009\)
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Tính giá trị của biểu thức:
\(N=\frac{\left(2^4+\frac{1}{4}\right).\left(4^4+\frac{1}{4}\right).\left(6^4+\frac{1}{4}\right)...\left(2008^4+\frac{1}{4}\right)}{\left(1^4+\frac{1}{4}\right).\left(3^4+\frac{1}{4}\right).\left(5^4+\frac{1}{4}\right)...\left(2007^4+\frac{1}{4}\right)}\)
Với mọi n thuộc N* ta có :
\(n^4+\frac{1}{4}=\left(n^4+2.\frac{1}{2}.n^2+\frac{1}{4}\right)-n^2=\left(n^2+\frac{1}{2}\right)^2-n^2\)
\(=\left(n^2+n+\frac{1}{2}\right)\left(n^2-n+\frac{1}{2}\right)\)
\(\Rightarrow N=\frac{\left(2^2+2+\frac{1}{2}\right)\left(2^2-2+\frac{1}{2}\right)...\left(2008^2+2008+\frac{1}{2}\right)\left(2008^2-2008+\frac{1}{2}\right)}{\left(1^2+1+\frac{1}{2}\right)\left(1^2-1+\frac{1}{2}\right)...\left(2007^2+2007+\frac{1}{2}\right)\left(2007^2-2007+\frac{1}{2}\right)}\)
\(=\frac{\left(2.3+\frac{1}{2}\right)\left(1.2+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right)...\left(2008.2009+\frac{1}{2}\right)}{\frac{1}{2}\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)...\left(2007.2008+\frac{1}{2}\right)}\)
\(=\frac{2008.2009+\frac{1}{2}}{\frac{1}{2}}=8068145\)
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