\(D=\left(\frac{1}{2}\right)^3+\left(\frac{1}{2}\right)^5+\left(\frac{1}{2}\right)^7+......+\left(\frac{1}{2}\right)^{2015}\)
a,\(\frac{-1}{24}-\left[\frac{1}{4}-\left(\frac{1}{2}-\frac{7}{8}\right)\right]\)
b,\(\left[\frac{5}{7}-\frac{7}{5}\right]-\left[\frac{1}{2}-\left(\frac{-2}{7}-\frac{1}{10}\right)\right]\)
c,\(\left(\frac{-1}{2}\right)-\left(\frac{-3}{5}\right)+\left(\frac{-1}{9}\right)+\frac{1}{71}-\left(\frac{-2}{7}\right)+\frac{4}{35}-\frac{7}{8}\)
d,\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
e,\(\left(\frac{1}{2}-\frac{13}{14}\right):\frac{5}{7}-\left(\frac{-2}{21}+\frac{1}{7}\right):\frac{5}{7}\)
g,\(\frac{4}{9}:\left(\frac{-1}{7}\right)+6\frac{5}{9}:\left(\frac{-1}{7}\right)\)
\(D=\frac{\left(2!\right)^2}{1^2}+\frac{\left(2!\right)^2}{3^2}+\frac{\left(2!\right)^2}{5^2}+\frac{\left(2!\right)^2}{7^2}+...+\frac{\left(2!\right)^2}{2015^2}\) so sánh D với 6
\(D=\frac{\left(2!\right)^2}{1^2}+\frac{\left(2!\right)^2}{3^2}+\frac{\left(2!\right)^2}{5^2}+\frac{\left(2!\right)^2}{7^2}+...+\frac{\left(2!\right)^2}{2015^2}\) so sánh D với 6
\(D=2!^2\left(\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{5^2}+...+\frac{1}{2015^2}\right)\)
tổng trong ngoặc nhỏ hơn 1 nên D nhỏ hơn 4.1=4<6
Vậy Đ<6
\(A=\left(6:\frac{3}{5}-1\frac{1}{6}x\frac{6}{7}\right):\left(4\frac{1}{5}x\frac{10}{11}+5\frac{2}{11}\right)\)\(B=\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{4}\right)x.......x\left(1-\frac{1}{2015}\right)x\left(1-\frac{1}{2016}\right)\)
\(C=5\frac{9}{10}:\frac{3}{2}-\left(2\frac{1}{3}x4\frac{1}{2}-2x2\frac{1}{3}\right):\frac{7}{4}\)
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:
\(D=\frac{4\cdot1}{2\cdot3}\cdot\frac{5\cdot2}{3\cdot4}\cdot\frac{6\cdot3}{4\cdot5}\cdot\frac{7\cdot4}{5\cdot6}\cdot\frac{8\cdot5}{6\cdot7}\cdot...\cdot\frac{41\cdot38}{39\cdot40}=\)
\(D=\frac{4\cdot5\cdot6\cdot7\cdot...\cdot41\times1\cdot2\cdot3\cdot4\cdot...\cdot38}{2\cdot3\cdot4\cdot5\cdot...\cdot39\times3\cdot4\cdot5\cdot6\cdot..\cdot40}=\frac{1}{39}\cdot\frac{41}{3}=\frac{41}{117}\)
\(\left(\frac{-5}{12}+\frac{7}{4}-\frac{3}{8}\right)-\left[4\frac{1}{2}-7\frac{1}{3}\right]-\left(\frac{1}{4}-\frac{5}{2}\right)\)
\(\left[2\frac{1}{4}-5\frac{3}{2}\right]-\left(\frac{3}{10}-1\right)-5\frac{1}{2}+\left(\frac{1}{3}-\frac{5}{6}\right)\)
\(\frac{4}{7}-\left(3\frac{2}{5}-1\frac{1}{2}\right)-\frac{5}{21}+\left[3\frac{1}{2}-4\frac{2}{3}\right]\)
\(\frac{1}{8}-1\frac{3}{4}+\left(\frac{7}{8}-3\frac{7}{2}+\frac{3}{4}\right)-\left[\frac{7}{4}-\frac{5}{8}\right]\)
\(\left(\frac{3}{5}-2\frac{1}{10}+\frac{11}{20}\right)-\left[\frac{-3}{4}+1\frac{7}{2}\right]\)
\(\left[-2\frac{1}{5}-2\frac{2}{3}\right]-\left(\frac{1}{15}-5\frac{1}{2}\right)+\left[\frac{-1}{6}+\frac{1}{3}\right]\)
\(1\frac{1}{8}-\left(\frac{1}{15}-\frac{1}{2}+\frac{-1}{6}\right)+\left[\frac{5}{4}+\frac{3}{2}\right]\)
\(\frac{5}{6}-\left(1\frac{1}{3}-1\frac{1}{2}\right)+\left[\frac{5}{12}-\frac{3}{4}-\frac{1}{6}\right]\)
\(1\frac{1}{4}-\left(\frac{7}{12}-\frac{2}{3}-1\frac{3}{8}\right)+\left[\frac{5}{24}-2\frac{1}{2}\right]-\frac{1}{6}-\left[\frac{-3}{4}\right]\)
\(-2\frac{1}{5}+2\frac{3}{10}-\left(\frac{6}{20}-\left[\frac{2}{8}-1\frac{1}{2}\right]\right)+\left[\frac{7}{20}-1\frac{1}{4}\right]\)
\(-\left[1\frac{2}{3}-3\frac{1}{2}+\frac{1}{4}\right]+\left(\frac{2}{6}-\frac{5}{12}\right)-\left(\frac{1}{3}-\left[\frac{1}{4}-\frac{1}{3}\right]\right)\)
\(-\frac{4}{5}-\left(1\frac{1}{10}-\frac{7}{10}\right)+\left[\frac{3}{4}-1\frac{1}{5}\right]+1\frac{1}{2}\)
\(\frac{3}{21}-\frac{5}{14}+\left[1\frac{1}{3}-5\frac{1}{2}+\frac{5}{14}\right]-\left(\frac{1}{6}-\frac{3}{7}+\frac{1}{3}\right)\)
\(-1\frac{2}{5}+\left[1\frac{3}{10}-\frac{7}{20}-1\frac{1}{4}\right]-\left(\frac{1}{5}-\left[\frac{3}{4}-1\frac{1}{2}\right]\right)\)
\(2\frac{1}{3}-\left(\frac{1}{2}-2\frac{1}{6}+\frac{3}{4}\right)+\left[\frac{5}{12}-1\frac{1}{3}\right]-\frac{7}{8}+3\frac{1}{2}\)
\(2\frac{1}{4}-1\frac{3}{5}-\left(\frac{9}{20}-\frac{7}{10}\right)+\left[1\frac{3}{5}-2\frac{1}{2}\right]+\frac{3}{4}\)
\(\left[\frac{8}{3}-5\frac{1}{4}+\frac{1}{6}\right]-\frac{7}{4}+\frac{-5}{12}-\left(1-1\frac{1}{2}+\frac{1}{3}\right)\)
\(\left(\frac{1}{4}-\left[1\frac{1}{4}-\frac{7}{10}\right]+\frac{1}{2}\right)-2\frac{1}{5}-1\frac{3}{10}+\left[1-\frac{1}{2}\right]\)
TRÌNH BÀY GIÚP MÌNH NHA
Bài 1 Thưc hiện phép tính ( tính nhanh nếu có thể)
a)\(\frac{-1}{24}-\left[\frac{1}{4}-\left(\frac{1}{2}-\frac{7}{8}\right)\right]\)
b)\(\left(\frac{5}{7}-\frac{7}{5}\right)-\left[\frac{1}{2}-\left(\frac{-2}{7}-\frac{1}{10}\right)\right]\)
C)\(\left(\frac{-1}{2}\right)-\left(\frac{-3}{5}\right)+\left(\frac{-1}{9}\right)+\frac{1}{17}-\left(\frac{-2}{7}\right)+\frac{4}{35}-\frac{7}{18}\)
d)\(\left(3-\frac{1}{4}+\frac{2}{3}\right)-\left(5-\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
Tính:
A=\(\left(1-\frac{2}{3}+\frac{4}{3}\right)-\left(\frac{4}{5}-1\right)+\left(\frac{7}{5}+2\right)\)
B=\(\left(-3+\frac{3}{4}-\frac{1}{3}\right):\left(5+\frac{2}{5}-\frac{2}{3}\right)\)
C=\(\left(\frac{3}{5}-\frac{4}{15}\right).\left(\frac{2}{7}-\frac{3}{14}\right)-\left(\frac{5}{9}-\frac{7}{27}\right)\)\(.\left(1-\frac{3}{5}\right)+\left(1-\frac{11}{12}\right).\left(1+\frac{11}{12}\right)\)
D=\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{-5}{19}}{\left(\frac{1}{14}+\frac{1}{7}-\frac{-3}{35}\right).\frac{4}{3}}\)
\(\left(1-\frac{2}{5}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{2}{2013}\right).\left(1-\frac{2}{2015}\right)\)
\(\left(1-\frac{2}{5}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{2}{2013}\right).\left(1-\frac{2}{2015}\right)\)
\(=\frac{3}{5}.\frac{5}{7}...\frac{2011}{2013}.\frac{2013}{2015}\)
\(=\frac{3.5...2011.2013}{5.7...2013.2015}\)
\(=\frac{3}{2015}\)
\(\left(1-\frac{2}{5}\right).\left(1-\frac{2}{7}\right)....\left(1-\frac{2}{2013}\right).\left(1-\frac{2}{2015}\right)\)
\(=\frac{3}{5}.\frac{5}{7}........\frac{2011}{2013}.\frac{2013}{2015}\)
\(=\frac{3}{2015}\)
\(\left(1-\frac{2}{5}\right).\left(1-\frac{2}{7}\right)...\left(1-\frac{2}{2015}\right)\)
\(=\frac{3}{5}.\frac{5}{7}...\frac{2013}{2015}\)
\(=\frac{3.5.7...2013}{5.7.9...2015}\)
\(=\frac{3}{2015}\)