B=\(\frac{5}{1\cdot4}\)+\(\frac{5}{4\cdot7}\)+\(\frac{5}{7\cdot10}\)+\(\frac{5}{10\cdot13}\)+\(\frac{5}{13\cdot16}\)
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}\)
\(\frac{15}{16}\)nha bạn
úm ba la xin tích
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+\frac{3}{13.16}\)
\(=1\left(\frac{1}{1}-\frac{1}{16}\right)\)
\(=1.\frac{15}{16}=\frac{15}{16}\)
3/1.4+3/4.7+3/7.10+3/10.13+3/13.16=1/1-1/4+1/4-1/7+1/7-1/10+1/10-1/13+1/13-1/16=1-1/16=15/15
\(\frac{5}{1\cdot4\cdot7}+\frac{5}{4\cdot7\cdot10}+\frac{5}{7\cdot10\cdot13}+.....+\frac{5}{31\cdot34\cdot37}\)Tính nhanh ( giải kiểu lớp 5 và dấu . là nhân)
THANKS nhiều
A=\(\frac{3}{1\cdot4}\)+\(\frac{3}{4\cdot7}\)+\(\frac{3}{7\cdot10}\)+\(\frac{3}{10\cdot13}\)+\(\frac{3}{13\cdot16}\)
\(A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}\)
\(A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}\)
\(A=1-\frac{1}{16}=\frac{15}{16}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{16}=1-\frac{1}{16}=\frac{15}{16}\)
\(\frac{2}{4\cdot7}-\frac{2}{5\cdot9}+\frac{2}{7\cdot10}-\frac{2}{9\cdot13}+\frac{2}{10\cdot13}-\frac{2}{13\cdot17}+...+\frac{2}{301\cdot304}-\frac{2}{401\cdot405}CM:tich,tren,< \frac{1}{15}\)
A=\(\frac{^{3^2}}{1\cdot4}\)+ \(\frac{3^2}{4\cdot7}\)+ \(\frac{3^2}{7\cdot10}\)+ \(\frac{3^2}{10\cdot13}\)+\(\frac{3^2}{13\cdot16}\)+......+ \(\frac{3^2}{97\cdot100}\)
A = \(\frac{3^2}{1\cdot4}+\frac{3^2}{4\cdot7}+\frac{3^2}{7\cdot10}+\frac{3^2}{10\cdot13}+\frac{3^2}{13\cdot16}+...+\frac{3^2}{97\cdot100}\)
A : 3 = \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+\frac{3}{13\cdot16}+...+\frac{3}{97\cdot100}\)
A : 3 = \(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+...+\frac{1}{97}-\frac{1}{100}\)
A : 3 = \(\frac{1}{1}-\frac{1}{100}\)
A : 3 = \(\frac{99}{100}\)
A = \(\frac{297}{100}\)
\(\frac{7}{3\cdot4}-\frac{9}{4\cdot5}+\frac{11}{5\cdot6}-\frac{13}{6\cdot7}+\frac{15}{7\cdot8}-\frac{17}{8\cdot9}+\frac{19}{9\cdot10}\)
ta có:(3+4)/4=3/(3*4)+4/(3*4) =1/4+1/3 chứng minh tương tự,cộng vế với vế, ta được kết quả là 13/30
Tính bằng phương pháp hợp lí
a> C = \(\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+\frac{3}{10\cdot13}+...+\frac{3}{73\cdot76}\) b> D = \(\frac{5}{10\cdot11}+\frac{5}{11\cdot12}+\frac{5}{12\cdot13}+...+\frac{5}{99\cdot100}\)
NHANH LÊN MÌNH CẦN GẤP
a) \(C=\frac{3}{4.7}+\frac{3}{7.10}+\frac{3}{10.13}+...+\frac{3}{73.76}\)
\(C=1.\left(\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+\frac{1}{10}-\frac{1}{13}+...+\frac{1}{73}-\frac{1}{76}\right)\)
\(C=1.\left(\frac{1}{4}-\frac{1}{76}\right)\)
\(C=1.\frac{9}{38}\)
\(C=\frac{9}{38}\)
b) \(D=\frac{5}{10.11}+\frac{5}{11.12}+\frac{5}{12.13}+...+\frac{5}{99.100}\)
\(D=5.\left(\frac{1}{10}-\frac{1}{11}+\frac{1}{11}-\frac{1}{12}+\frac{1}{12}-\frac{1}{13}+...+\frac{1}{99}+\frac{1}{100}\right)\)
\(D=5.\left(\frac{1}{10}-\frac{1}{100}\right)\)
\(D=5.\frac{9}{100}\)
\(D=\frac{99}{20}\)
Tính bằng cách hợp lý (nếu có thể) :
A = \(\frac{6}{2\cdot2}+\frac{5}{2\cdot13}+\frac{3}{13\cdot4}+\frac{2}{4\cdot18}+\frac{10}{18\cdot7}\)
\(\Rightarrow A=4.\left[\frac{6}{2.\left(2.4\right)}+\frac{5}{\left(2.4\right).13}+\frac{3}{13.\left(4.4\right)}+\frac{2}{\left(4.4\right).18}+\frac{10}{18.\left(7.4\right)}\right]\)
\(=4.\left(\frac{6}{2.8}+\frac{5}{8.13}+\frac{3}{13.16}+\frac{2}{16.18}+\frac{10}{18.28}\right)=4.\left(\frac{1}{2}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+\frac{1}{13}-\frac{1}{16}+\frac{1}{16}-\frac{1}{18}+\frac{1}{18}-\frac{1}{28}\right)\)
\(=4.\left(\frac{1}{2}-\frac{1}{28}\right)=4.\frac{13}{28}=\frac{13}{7}\)
Chứng tỏ rằng :
\(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+\frac{1}{10\cdot13}+...\frac{1}{2010\cdot2013}<\frac{1}{3}\)
\(=\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+....+\frac{3}{2010.2013}\right)\)
\(=\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{2010}-\frac{1}{2013}\right)\)
\(=\frac{1}{3}\left(1-\frac{1}{2013}\right)=\frac{1}{3}.\frac{2012}{2013}